\displaystyle \textbf{Question 1: }\text{If two isosceles triangles have a common base, prove that the line joining}
\displaystyle \text{their vertices bisects the common base at right angles.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }ABC\text{ and }DBC\text{ be two isosceles triangles having the common base }BC.
\displaystyle \text{Therefore, }AB=AC\text{ and }DB=DC.2018-09-02_10-06-13\displaystyle \text{Join }AD,\text{ and let it meet }BC\text{ at }E.
\displaystyle \text{We have to prove that }BE=CE\text{ and }AD\perp BC.
\displaystyle \text{In }\triangle ABD\text{ and }\triangle ACD,
\displaystyle AB=AC\qquad\text{(Given)}
\displaystyle BD=CD\qquad\text{(Given)}
\displaystyle AD=AD\qquad\text{(Common)}
\displaystyle \therefore \triangle ABD\cong\triangle ACD\qquad\text{(SSS congruence criterion)}
\displaystyle \therefore \angle BAD=\angle DAC\qquad\text{(Corresponding parts of congruent triangles)}
\displaystyle \text{Since }E\text{ lies on }AD,
\displaystyle \angle BAE=\angle EAC.
\displaystyle \text{In }\triangle ABE\text{ and }\triangle ACE,
\displaystyle AB=AC\qquad\text{(Given)}
\displaystyle AE=AE\qquad\text{(Common)}
\displaystyle \angle BAE=\angle EAC
\displaystyle \therefore \triangle ABE\cong\triangle ACE\qquad\text{(SAS congruence criterion)}
\displaystyle \therefore BE=CE\text{ and }\angle AEB=\angle AEC.
\displaystyle \text{Since }B,E,C\text{ are collinear,}
\displaystyle \angle AEB+\angle AEC=180^\circ.
\displaystyle \text{Also, }\angle AEB=\angle AEC.
\displaystyle \therefore \angle AEB=\angle AEC=90^\circ.
\displaystyle \therefore BE=CE\text{ and }AD\perp BC.
\displaystyle \therefore AD\text{ bisects }BC\text{ at right angles.}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\triangle ABC\text{ and }\triangle DBC\text{ are two isosceles triangles on the same base }BC,
\displaystyle \text{and their vertices }A\text{ and }D\text{ lie on the same side of }BC.\text{ If }AD\text{ produced meets}
\displaystyle BC\text{ at }E,\text{ show that:}
\displaystyle \text{(i) }\triangle ABD\cong\triangle ACD\qquad\text{(ii) }\triangle ABE\cong\triangle ACE
\displaystyle \text{(iii) }AE\text{ bisects }\angle BAC\text{ as well as }\angle BDC.
\displaystyle \text{Answer:} 2018-09-02_10-06-38
\displaystyle \text{Since }\triangle ABC\text{ and }\triangle DBC\text{ are isosceles triangles,}
\displaystyle AB=AC\quad\text{and}\quad DB=DC.$

\displaystyle \text{(i) In }\triangle ABD\text{ and }\triangle ACD,
\displaystyle AB=AC\qquad\text{(Given)}
\displaystyle BD=CD\qquad\text{(Given)}
\displaystyle AD=AD\qquad\text{(Common)}
\displaystyle \therefore \triangle ABD\cong\triangle ACD\qquad\text{(SSS congruence criterion)}

\displaystyle \text{(ii) From part (i),}
\displaystyle \angle BAD=\angle DAC\qquad\text{(Corresponding parts of congruent triangles)}
\displaystyle \text{Since }A,D,E\text{ are collinear,}
\displaystyle \angle BAE=\angle EAC.
\displaystyle \text{In }\triangle ABE\text{ and }\triangle ACE,
\displaystyle AB=AC\qquad\text{(Given)}
\displaystyle AE=AE\qquad\text{(Common)}
\displaystyle \angle BAE=\angle EAC
\displaystyle \therefore \triangle ABE\cong\triangle ACE\qquad\text{(SAS congruence criterion)}

\displaystyle \text{(iii) From }\triangle ABD\cong\triangle ACD,
\displaystyle \angle BAD=\angle DAC.
\displaystyle \text{Since }A,D,E\text{ are collinear,}
\displaystyle \angle BAE=\angle EAC.
\displaystyle \therefore AE\text{ bisects }\angle BAC.
\displaystyle \text{Also, from }\triangle ABD\cong\triangle ACD,
\displaystyle \angle BDA=\angle ADC.
\displaystyle \text{Since }A,D,E\text{ are collinear,}
\displaystyle \angle BDE=\angle EDC.
\displaystyle \therefore DE,\text{ and hence }AE,\text{ bisects }\angle BDC.
\displaystyle \therefore AE\text{ bisects }\angle BAC\text{ as well as }\angle BDC.
\displaystyle \therefore \text{All the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{A point }E\text{ is taken inside a rhombus }ABCD\text{ such that its distances from}
\displaystyle \text{the vertices }D\text{ and }B\text{ are equal. Show that }AE\text{ and }EC\text{ lie on the same}
\displaystyle \text{straight line.}
\displaystyle \text{Answer:} 2018-09-02_10-06-49
\displaystyle \text{Given }BE=DE.
\displaystyle \text{Since }ABCD\text{ is a rhombus,}
\displaystyle AB=BC=CD=DA.
\displaystyle \text{In }\triangle AED\text{ and }\triangle AEB,
\displaystyle AD=AB\qquad\text{(Sides of a rhombus)}
\displaystyle DE=BE\qquad\text{(Given)}
\displaystyle AE=AE\qquad\text{(Common)}
\displaystyle \therefore \triangle AED\cong\triangle AEB\qquad\text{(SSS congruence criterion)}
\displaystyle \therefore \angle AED=\angle AEB\qquad\text{(Corresponding parts of congruent triangles)}
\displaystyle \text{In }\triangle DEC\text{ and }\triangle BEC,
\displaystyle DC=BC\qquad\text{(Sides of a rhombus)}
\displaystyle DE=BE\qquad\text{(Given)}
\displaystyle EC=EC\qquad\text{(Common)}
\displaystyle \therefore \triangle DEC\cong\triangle BEC\qquad\text{(SSS congruence criterion)}
\displaystyle \therefore \angle DEC=\angle BEC\qquad\text{(Corresponding parts of congruent triangles)}
\displaystyle \text{The sum of the angles around the point }E\text{ is }360^\circ.
\displaystyle \therefore \angle AED+\angle DEC+\angle BEC+\angle AEB=360^\circ
\displaystyle 2\angle AED+2\angle DEC=360^\circ
\displaystyle \therefore \angle AED+\angle DEC=180^\circ
\displaystyle \therefore \angle AEC=180^\circ.
\displaystyle \therefore A,\ E\text{ and }C\text{ are collinear.}
\displaystyle \therefore AE\text{ and }EC\text{ lie on the same straight line.}
\displaystyle \text{Hence, the required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In the adjoining figure, it is given that }AB=CD\text{ and }AC=BD.
\displaystyle \text{Prove that }\triangle ADC\cong\triangle DAB. 2018-09-02_10-07-02\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ADC\text{ and }\triangle DAB,
\displaystyle AC=DB\qquad\text{(Given)}
\displaystyle CD=AB\qquad\text{(Given)}
\displaystyle AD=DA\qquad\text{(Common)}
\displaystyle \therefore \triangle ADC\cong\triangle DAB\qquad\text{(SSS congruence criterion)}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{In }\triangle ABC,\text{ if }AB=AC\text{ and }D,\ F\text{ and }E\text{ are the}
\displaystyle \text{midpoints of the sides }AB,\ BC\text{ and }AC,\text{ respectively, prove that }DF=EF.
\displaystyle \text{Answer:} 2018-09-02_10-07-19\displaystyle \text{Since }D,\ F\text{ and }E\text{ are the midpoints of }AB,\ BC\text{ and }AC,\text{ respectively,}
\displaystyle AD=DB,\quad BF=FC\quad\text{and}\quad AE=EC.
\displaystyle \text{Also, }AB=AC\text{ (Given).}
\displaystyle \therefore AD=AE=\frac12AB=\frac12AC.
\displaystyle \text{In }\triangle ABC,\text{ }D\text{ and }F\text{ are the midpoints of }AB\text{ and }BC.
\displaystyle \therefore DF\parallel AC\quad\text{and}\quad DF=\frac12AC\qquad\text{(Midpoint Theorem)}
\displaystyle \text{Similarly, in }\triangle ABC,\text{ }E\text{ and }F\text{ are the midpoints of }AC\text{ and }BC.
\displaystyle \therefore EF\parallel AB\quad\text{and}\quad EF=\frac12AB\qquad\text{(Midpoint Theorem)}
\displaystyle \text{Since }AB=AC,
\displaystyle \frac12AB=\frac12AC.
\displaystyle \therefore EF=DF.
\displaystyle \therefore DF=EF.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{In the adjoining figure, it is given that }DF=FE,\ BF=FC,\ FD\perp AB
\displaystyle \text{and }FE\perp AC.\text{ Prove that }AB=AC.\text{ Hence, prove that }\triangle ABC\text{ is isosceles.} 2018-09-02_10-07-33\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle BDF\text{ and }\triangle CEF,
\displaystyle BF=CF\qquad\text{(Given)}
\displaystyle DF=EF\qquad\text{(Given)}
\displaystyle \angle BDF=\angle CEF=90^\circ
\displaystyle \therefore \triangle BDF\cong\triangle CEF\qquad\text{(RHS congruence criterion)}
\displaystyle \therefore \angle DBF=\angle ECF\qquad\text{(Corresponding parts of congruent triangles)}
\displaystyle \text{Since }D\text{ lies on }AB,\ F\text{ lies on }BC\text{ and }E\text{ lies on }AC,
\displaystyle \angle DBF=\angle ABC\quad\text{and}\quad\angle ECF=\angle BCA.
\displaystyle \therefore \angle ABC=\angle BCA.
\displaystyle \therefore AB=AC\qquad\text{(Sides opposite equal angles are equal)}
\displaystyle \therefore \triangle ABC\text{ is an isosceles triangle.}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }ABCD\text{ is a square. }E\text{ and }F\text{ are points on the sides }AD\text{ and }BC,
\displaystyle \text{respectively, such that }AF=BE.\text{ Prove that }BF=AE\text{ and }\angle BAF=\angle ABE.
\displaystyle \text{Answer:} 2018-09-02_10-07-45\displaystyle \text{In }\triangle ABE\text{ and }\triangle BAF,
\displaystyle BE=AF\qquad\text{(Given)}
\displaystyle AB=BA\qquad\text{(Common)}
\displaystyle \text{Since }ABCD\text{ is a square, }AB\perp AD\text{ and }AB\perp BC.
\displaystyle \therefore \angle BAE=\angle ABF=90^\circ.
\displaystyle \therefore \triangle ABE\cong\triangle BAF\qquad\text{(RHS congruence criterion)}
\displaystyle \therefore AE=BF\qquad\text{(Corresponding parts of congruent triangles)}
\displaystyle \text{Also, }\angle ABE=\angle BAF\qquad\text{(Corresponding parts of congruent triangles)}
\displaystyle \therefore BF=AE\text{ and }\angle BAF=\angle ABE.
\displaystyle \therefore \text{The required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the adjoining figure, }AB>AC.\ BD\text{ and }CD\text{ are the bisectors of}
\displaystyle \angle ABC\text{ and }\angle ACB,\text{ respectively. Prove that }DB>DC. 2018-09-02_10-08-15\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,
\displaystyle AB>AC\qquad\text{(Given)}
\displaystyle \therefore \angle ACB>\angle ABC
\displaystyle \text{(The angle opposite the greater side of a triangle is greater.)}
\displaystyle \therefore \frac12\angle ACB>\frac12\angle ABC.
\displaystyle \text{Since }CD\text{ bisects }\angle ACB,
\displaystyle \angle DCB=\frac12\angle ACB.
\displaystyle \text{Since }BD\text{ bisects }\angle ABC,
\displaystyle \angle DBC=\frac12\angle ABC.
\displaystyle \therefore \angle DCB>\angle DBC.
\displaystyle \text{In }\triangle BDC,\text{ the side opposite the greater angle is greater.}
\displaystyle \therefore DB>DC.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the adjoining figure, the sides }AB\text{ and }AC\text{ of }\triangle ABC\text{ are}
\displaystyle \text{extended to }D\text{ and }E,\text{ respectively. If }x>y,\text{ show that }AB>AC. 2018-09-02_10-08-28\displaystyle \text{Answer:}
\displaystyle \text{Since }\angle ABC\text{ and }x\text{ form a linear pair,}
\displaystyle \angle ABC+x=180^\circ
\displaystyle \therefore \angle ABC=180^\circ-x\qquad\ldots\text{(i)}
\displaystyle \text{Similarly, }\angle ACB\text{ and }y\text{ form a linear pair.}
\displaystyle \angle ACB+y=180^\circ
\displaystyle \therefore \angle ACB=180^\circ-y\qquad\ldots\text{(ii)}
\displaystyle \text{Given that }x>y,
\displaystyle 180^\circ-x<180^\circ-y
\displaystyle \therefore \angle ABC<\angle ACB\qquad\text{[Using (i) and (ii)]}
\displaystyle \text{In }\triangle ABC,\text{ the side opposite the greater angle is greater.}
\displaystyle \therefore AB>AC.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }D\text{ is any point on the base }BC\text{ produced, of an isosceles}
\displaystyle \triangle ABC,\text{ prove that }AD>AB.
\displaystyle \text{Answer:}
\displaystyle \text{Since }\triangle ABC\text{ is isosceles,}
\displaystyle AB=AC.
\displaystyle \therefore \angle ABC=\angle ACB\qquad\text{(Angles opposite equal sides are equal)}\qquad\ldots\text{(i)}
\displaystyle \text{In }\triangle ABD,\text{ }\angle ACD\text{ is an exterior angle.}
\displaystyle \therefore \angle ACD>\angle ABD\qquad\text{(Exterior angle theorem)}\qquad\ldots\text{(ii)}
\displaystyle \text{Since }B,\ C\text{ and }D\text{ are collinear,}2018-09-02_10-08-45
\displaystyle \angle ACD=180^\circ-\angle ACB
\displaystyle \text{and }\angle ABD=\angle ABC.
\displaystyle \text{From (i) and (ii),}
\displaystyle 180^\circ-\angle ACB>\angle ACB
\displaystyle \therefore \angle ACD>\angle ACB.
\displaystyle \text{In }\triangle ACD,\text{ the side opposite the greater angle is greater.}
\displaystyle \therefore AD>AC.
\displaystyle \text{But }AB=AC.
\displaystyle \therefore AD>AB.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the adjoining figure, if }AD\text{ is the bisector of }\angle BAC,\text{ show that:}
\displaystyle \text{(i) }AB>BD\qquad\text{(ii) }AC>CD. 2018-09-02_10-09-08\displaystyle \text{Answer:}
\displaystyle \text{Since }AD\text{ bisects }\angle BAC,
\displaystyle \angle BAD=\angle DAC\qquad\ldots\text{(1)}

\displaystyle \text{(i) In }\triangle ADC,\ \angle BDA\text{ is an exterior angle.}
\displaystyle \therefore \angle BDA>\angle DAC
\displaystyle \text{(An exterior angle of a triangle is greater than either interior opposite angle.)}
\displaystyle \therefore \angle BDA>\angle BAD\qquad\text{[Using (1)]}
\displaystyle \text{In }\triangle ABD,\text{ the side opposite the greater angle is greater.}
\displaystyle \therefore AB>BD.

\displaystyle \text{(ii) In }\triangle ABD,\ \angle ADC\text{ is an exterior angle.}
\displaystyle \therefore \angle ADC>\angle BAD
\displaystyle \text{(An exterior angle of a triangle is greater than either interior opposite angle.)}
\displaystyle \therefore \angle ADC>\angle DAC\qquad\text{[Using (1)]}
\displaystyle \text{In }\triangle ACD,\text{ the side opposite the greater angle is greater.}
\displaystyle \therefore AC>CD.
\displaystyle \therefore \text{Both the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the adjoining figure, }AC>AB\text{ and }AD\text{ is the bisector of}
\displaystyle \angle BAC.\text{ Show that }\angle ADC>\angle ADB. 2018-09-02_10-09-24\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,\ AC>AB.
\displaystyle \therefore \angle ABC>\angle ACB
\displaystyle \text{(The angle opposite the greater side of a triangle is greater.)}\qquad\ldots\text{(i)}
\displaystyle \text{Since }AD\text{ bisects }\angle BAC,
\displaystyle \angle BAD=\angle DAC\qquad\ldots\text{(ii)}
\displaystyle \text{In }\triangle ABD,
\displaystyle \angle ADB=180^\circ-\angle ABD-\angle BAD
\displaystyle =180^\circ-\angle ABC-\angle BAD\qquad\ldots\text{(iii)}
\displaystyle \text{In }\triangle ACD,
\displaystyle \angle ADC=180^\circ-\angle ACD-\angle DAC
\displaystyle =180^\circ-\angle ACB-\angle DAC\qquad\ldots\text{(iv)}
\displaystyle \text{From (i) and (ii),}
\displaystyle \angle ABC+\angle BAD>\angle ACB+\angle DAC.
\displaystyle \therefore 180^\circ-(\angle ABC+\angle BAD)
\displaystyle <180^\circ-(\angle ACB+\angle DAC)
\displaystyle \therefore \angle ADB<\angle ADC\qquad\text{[Using (iii) and (iv)]}
\displaystyle \therefore \angle ADC>\angle ADB.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Show that the sum of the three altitudes of a triangle is less than the}
\displaystyle \text{sum of the three sides of the triangle.}
\displaystyle \text{Answer:} 2018-09-02_10-09-36\displaystyle \text{Let }AD,\ BE\text{ and }CF\text{ be the altitudes of }\triangle ABC.
\displaystyle \therefore AD\perp BC,\quad BE\perp AC,\quad CF\perp AB.
\displaystyle \text{In right-angled }\triangle ADC,\ AC>AD,
\displaystyle \text{and in right-angled }\triangle ADB,\ AB>AD.
\displaystyle \therefore AB+AC>2AD\qquad\ldots\text{(i)}
\displaystyle \text{In right-angled }\triangle ABE,\ AB>BE,
\displaystyle \text{and in right-angled }\triangle BEC,\ BC>BE.
\displaystyle \therefore AB+BC>2BE\qquad\ldots\text{(ii)}
\displaystyle \text{In right-angled }\triangle CFA,\ AC>CF,
\displaystyle \text{and in right-angled }\triangle CFB,\ BC>CF.
\displaystyle \therefore AC+BC>2CF\qquad\ldots\text{(iii)}
\displaystyle \text{Adding (i), (ii) and (iii), we get}
\displaystyle (AB+AC)+(AB+BC)+(AC+BC)>2AD+2BE+2CF
\displaystyle 2(AB+BC+AC)>2(AD+BE+CF)
\displaystyle \therefore AB+BC+AC>AD+BE+CF.
\displaystyle \therefore \text{The sum of the three altitudes is less than the sum of the three sides of the triangle.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Prove that the perimeter of a triangle is greater than the sum of its}
\displaystyle \text{three medians.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }AD,\ BE\text{ and }CF\text{ be the medians of }\triangle ABC.
\displaystyle \text{We know that the sum of any two sides of a triangle is greater than twice the}
\displaystyle \text{median drawn to the third side.}2018-09-02_10-09-57
\displaystyle \text{Since }AD\text{ is the median drawn to }BC,
\displaystyle AB+AC>2AD\qquad\ldots\text{(i)}
\displaystyle \text{Since }BE\text{ is the median drawn to }AC,
\displaystyle AB+BC>2BE\qquad\ldots\text{(ii)}
\displaystyle \text{Since }CF\text{ is the median drawn to }AB,
\displaystyle AC+BC>2CF\qquad\ldots\text{(iii)}
\displaystyle \text{Adding (i), (ii) and (iii), we get}
\displaystyle (AB+AC)+(AB+BC)+(AC+BC)>2AD+2BE+2CF
\displaystyle 2(AB+BC+AC)>2(AD+BE+CF)
\displaystyle \therefore AB+BC+AC>AD+BE+CF.
\displaystyle \therefore \text{The perimeter of the triangle is greater than the sum of its three medians.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the adjoining figure, }ABC\text{ is a triangle and }D\text{ is any point in its}
\displaystyle \text{interior. Show that }DB+DC<AB+AC. 2018-09-02_10-10-10\displaystyle \text{Answer:}
\displaystyle \text{Let }BD\text{ produced meet }AC\text{ at }E,\text{ as shown in the figure.}
\displaystyle \text{In }\triangle ABE,
\displaystyle AB+AE>BE\qquad\text{(The sum of any two sides of a triangle is greater than the third side.)}
\displaystyle \text{Since }B,\ D\text{ and }E\text{ are collinear,}
\displaystyle BE=BD+DE.
\displaystyle \therefore AB+AE>BD+DE\qquad\ldots\text{(i)}
\displaystyle \text{In }\triangle CDE,
\displaystyle DE+EC>DC\qquad\text{(The sum of any two sides of a triangle is greater than the third side.)}\qquad\ldots\text{(ii)}
\displaystyle \text{Adding (i) and (ii), we get}
\displaystyle AB+AE+DE+EC>BD+DE+DC
\displaystyle \therefore AB+(AE+EC)>BD+DC
\displaystyle \text{Since }A,\ E\text{ and }C\text{ are collinear,}
\displaystyle AE+EC=AC.
\displaystyle \therefore AB+AC>BD+DC.
\displaystyle \therefore DB+DC<AB+AC.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the adjoining figure, }ABCD\text{ is a quadrilateral. }AB\text{ is the longest}
\displaystyle \text{side and }CD\text{ is the shortest side. Prove that }\angle DCB>\angle DAB\text{ and}
\displaystyle \angle ADC>\angle ABC. 2018-09-02_10-10-30\displaystyle \text{Answer:}
\displaystyle \text{Join }AC\text{ and }BD.
\displaystyle \text{(i) Since }AB\text{ is the longest side of quadrilateral }ABCD,
\displaystyle AB>BC.
\displaystyle \text{Therefore, in }\triangle ABC,
\displaystyle \angle ACB>\angle CAB\qquad\ldots\text{(1)}
\displaystyle \text{Also, since }CD\text{ is the shortest side of quadrilateral }ABCD,
\displaystyle AD>CD.
\displaystyle \text{Therefore, in }\triangle ACD,
\displaystyle \angle ACD>\angle CAD\qquad\ldots\text{(2)}
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle \angle ACB+\angle ACD>\angle CAB+\angle CAD
\displaystyle \therefore \angle DCB>\angle DAB.
\displaystyle \text{(ii) Since }AB\text{ is the longest side of quadrilateral }ABCD,
\displaystyle AB>AD.
\displaystyle \text{Therefore, in }\triangle ABD,
\displaystyle \angle ADB>\angle DBA\qquad\ldots\text{(3)}
\displaystyle \text{Also, since }CD\text{ is the shortest side of quadrilateral }ABCD,
\displaystyle BC>CD.
\displaystyle \text{Therefore, in }\triangle BCD,
\displaystyle \angle BDC>\angle DBC\qquad\ldots\text{(4)}
\displaystyle \text{Adding (3) and (4), we get}
\displaystyle \angle ADB+\angle BDC>\angle DBA+\angle DBC
\displaystyle \therefore \angle ADC>\angle ABC.
\displaystyle \therefore \text{The required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In the adjoining figure, }ABCD\text{ is a quadrilateral whose diagonals }AC
\displaystyle \text{and }BD\text{ intersect at }O.\text{ Show that:}
\displaystyle \text{(i) }AB+BC+CD+DA>AC+BD
\displaystyle \text{(ii) }AB+BC+CD+DA<2(AC+BD). 2018-09-02_10-10-41\displaystyle \text{Answer:}
\displaystyle \text{(i) In }\triangle ABC,\text{ by the triangle inequality,}
\displaystyle AB+BC>AC\qquad\ldots\text{(1)}
\displaystyle \text{In }\triangle ADC,
\displaystyle AD+DC>AC\qquad\ldots\text{(2)}
\displaystyle \text{In }\triangle ABD,
\displaystyle AB+AD>BD\qquad\ldots\text{(3)}
\displaystyle \text{In }\triangle BCD,
\displaystyle BC+CD>BD\qquad\ldots\text{(4)}
\displaystyle \text{Adding (1), (2), (3) and (4), we get}
\displaystyle 2(AB+BC+CD+DA)>2(AC+BD)
\displaystyle \therefore AB+BC+CD+DA>AC+BD.

\displaystyle \text{(ii) In }\triangle OAB,\text{ by the triangle inequality,}
\displaystyle OA+OB>AB\qquad\ldots\text{(5)}
\displaystyle \text{In }\triangle OBC,
\displaystyle OB+OC>BC\qquad\ldots\text{(6)}
\displaystyle \text{In }\triangle OCD,
\displaystyle OC+OD>CD\qquad\ldots\text{(7)}
\displaystyle \text{In }\triangle ODA,
\displaystyle OD+OA>DA\qquad\ldots\text{(8)}
\displaystyle \text{Adding (5), (6), (7) and (8), we get}
\displaystyle 2(OA+OB+OC+OD)>AB+BC+CD+DA
\displaystyle \text{Since }O\text{ lies on }AC\text{ and }BD,
\displaystyle OA+OC=AC\quad\text{and}\quad OB+OD=BD.
\displaystyle \therefore 2(AC+BD)>AB+BC+CD+DA
\displaystyle \therefore AB+BC+CD+DA<2(AC+BD).
\displaystyle \therefore \text{Both the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }O\text{ is any point in the interior of }\triangle ABC.\text{ Prove that:}
\displaystyle \text{(i) }AB+AC>OB+OC
\displaystyle \text{(ii) }AB+BC+CA>OA+OB+OC
\displaystyle \text{(iii) }OA+OB+OC>\frac12(AB+BC+CA).
\displaystyle \text{Answer:} 2018-09-02_10-10-56
\displaystyle \text{(i) Let }BO\text{ produced meet }AC\text{ at }D.
\displaystyle \text{In }\triangle ABD,\text{ by the triangle inequality,}
\displaystyle AB+AD>BD
\displaystyle \text{Since }B,\ O\text{ and }D\text{ are collinear, }BD=BO+OD.
\displaystyle \therefore AB+AD>BO+OD\qquad\ldots\text{(1)}
\displaystyle \text{In }\triangle ODC,\text{ by the triangle inequality,}
\displaystyle OD+DC>OC\qquad\ldots\text{(2)}
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle AB+AD+OD+DC>BO+OD+OC
\displaystyle \therefore AB+(AD+DC)>BO+OC
\displaystyle \text{Since }A,\ D\text{ and }C\text{ are collinear, }AD+DC=AC.
\displaystyle \therefore AB+AC>OB+OC.

\displaystyle \text{(ii) By proceeding similarly with }CO\text{ and }AO\text{ produced, we obtain}
\displaystyle BC+BA>OC+OA\qquad\ldots\text{(3)}
\displaystyle CA+CB>OA+OB\qquad\ldots\text{(4)}
\displaystyle \text{From part (i),}
\displaystyle AB+AC>OB+OC\qquad\ldots\text{(5)}
\displaystyle \text{Adding (3), (4) and (5), we get}
\displaystyle 2(AB+BC+CA)>2(OA+OB+OC)
\displaystyle \therefore AB+BC+CA>OA+OB+OC.

\displaystyle \text{(iii) In }\triangle OAB,\text{ by the triangle inequality,}
\displaystyle OA+OB>AB\qquad\ldots\text{(6)}
\displaystyle \text{In }\triangle OBC,
\displaystyle OB+OC>BC\qquad\ldots\text{(7)}
\displaystyle \text{In }\triangle OCA,
\displaystyle OC+OA>CA\qquad\ldots\text{(8)}
\displaystyle \text{Adding (6), (7) and (8), we get}
\displaystyle 2(OA+OB+OC)>AB+BC+CA
\displaystyle \therefore OA+OB+OC>\frac12(AB+BC+CA).
\displaystyle \therefore \text{All the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{In the adjoining figure, prove that:}
\displaystyle \text{(i) }CD+DA+AB+BC>2AC\qquad\text{(ii) }CD+DA+AB>BC. 2018-09-02_10-11-06\displaystyle \text{Answer:}
\displaystyle \text{(i) In }\triangle ABC,\text{ by the triangle inequality,}
\displaystyle AB+BC>AC\qquad\ldots\text{(1)}
\displaystyle \text{In }\triangle ACD,\text{ by the triangle inequality,}
\displaystyle AD+CD>AC\qquad\ldots\text{(2)}
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle AB+BC+AD+CD>2AC.
\displaystyle \therefore CD+DA+AB+BC>2AC.

\displaystyle \text{(ii) In }\triangle ACD,\text{ by the triangle inequality,}
\displaystyle CD+DA>CA.
\displaystyle \therefore CD+DA+AB>CA+AB\qquad\ldots\text{(3)}
\displaystyle \text{In }\triangle ABC,\text{ by the triangle inequality,}
\displaystyle CA+AB>BC\qquad\ldots\text{(4)}
\displaystyle \text{From (3) and (4),}
\displaystyle \therefore CD+DA+AB>BC.
\displaystyle \therefore \text{Both the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In the adjoining figure, }AB=AC\text{ and }DB=DC.\text{ Prove that}
\displaystyle \angle ABD=\angle ACD. 2018-08-11_8-38-23\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,
\displaystyle AB=AC\qquad\text{(Given)}
\displaystyle \therefore \angle ABC=\angle ACB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}\qquad\ldots\text{(1)}
\displaystyle \text{In }\triangle DBC,
\displaystyle DB=DC\qquad\text{(Given)}
\displaystyle \therefore \angle DBC=\angle DCB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}\qquad\ldots\text{(2)}
\displaystyle \text{Since }BD\text{ lies inside }\angle ABC,
\displaystyle \angle ABD=\angle ABC-\angle DBC.
\displaystyle \text{Similarly, since }CD\text{ lies inside }\angle ACB,
\displaystyle \angle ACD=\angle ACB-\angle DCB.
\displaystyle \text{Using (1) and (2),}
\displaystyle \angle ABC-\angle DBC=\angle ACB-\angle DCB.
\displaystyle \therefore \angle ABD=\angle ACD.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In the adjoining figure, }AB=AC.\ BE\text{ and }CF\text{ are the bisectors of}
\displaystyle \angle ABC\text{ and }\angle ACB,\text{ respectively. Prove that }\triangle EBC\cong\triangle FCB. 2018-08-11_8-38-36\displaystyle \text{Answer:}
\displaystyle \text{Since }AB=AC,
\displaystyle \angle ABC=\angle ACB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}\qquad\ldots\text{(1)}
\displaystyle \text{Since }BE\text{ bisects }\angle ABC,
\displaystyle \angle EBC=\frac12\angle ABC.
\displaystyle \text{Since }CF\text{ bisects }\angle ACB,
\displaystyle \angle FCB=\frac12\angle ACB.
\displaystyle \therefore \angle EBC=\angle FCB\qquad\text{[Using (1)]}
\displaystyle \text{Also, since }F\text{ lies on }AB,
\displaystyle \angle FBC=\angle ABC.
\displaystyle \text{Since }E\text{ lies on }AC,
\displaystyle \angle ECB=\angle ACB.
\displaystyle \therefore \angle FBC=\angle ECB\qquad\text{[Using (1)]}
\displaystyle \text{In }\triangle EBC\text{ and }\triangle FCB,
\displaystyle \angle EBC=\angle FCB
\displaystyle BC=CB\qquad\text{(Common)}
\displaystyle \angle ECB=\angle FBC
\displaystyle \therefore \triangle EBC\cong\triangle FCB\qquad\text{(ASA congruence criterion)}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{If }\triangle ABC\text{ is an isosceles triangle with }AB=AC,\text{ prove that the}
\displaystyle \text{perpendiculars drawn from the vertices }B\text{ and }C\text{ to their opposite sides are equal.}
\displaystyle \text{Answer:} 2018-08-11_8-38-52\displaystyle \text{Let }BD\perp AC\text{ and }CE\perp AB,\text{ where }D\text{ lies on }AC\text{ and }E\text{ lies on }AB.
\displaystyle \text{Since }AB=AC,
\displaystyle \angle ABC=\angle ACB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}
\displaystyle \text{Since }E\text{ lies on }AB\text{ and }D\text{ lies on }AC,
\displaystyle \angle EBC=\angle ABC\quad\text{and}\quad\angle DCB=\angle ACB.
\displaystyle \therefore \angle EBC=\angle DCB.
\displaystyle \text{In }\triangle EBC\text{ and }\triangle DCB,
\displaystyle \angle BEC=\angle CDB=90^\circ
\displaystyle \angle EBC=\angle DCB
\displaystyle BC=CB\qquad\text{(Common)}
\displaystyle \therefore \triangle EBC\cong\triangle DCB\qquad\text{(AAS congruence criterion)}
\displaystyle \therefore CE=BD\qquad\text{(Corresponding parts of congruent triangles)}
\displaystyle \therefore \text{The perpendiculars drawn from }B\text{ and }C\text{ to their opposite sides are equal.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{In the adjoining figure, it is given that }\angle BAD=\angle BCE\text{ and}
\displaystyle AB=BC.\text{ Prove that }\triangle ABD\cong\triangle CBE. 2018-08-11_8-39-04\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABD\text{ and }\triangle CBE,
\displaystyle \angle BAD=\angle BCE\qquad\text{(Given)}
\displaystyle \text{Since }D\text{ lies on }BC,\ \angle ABD=\angle ABC.
\displaystyle \text{Since }E\text{ lies on }BA,\ \angle CBE=\angle CBA.
\displaystyle \text{But }\angle ABC=\angle CBA.
\displaystyle \therefore \angle ABD=\angle CBE.
\displaystyle AB=BC\qquad\text{(Given)}
\displaystyle \therefore \triangle ABD\cong\triangle CBE\qquad\text{(ASA congruence criterion)}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{In }\triangle ABC,\ AB=AC,\text{ and the bisectors of }\angle ABC\text{ and}
\displaystyle \angle ACB\text{ intersect at }O.\text{ Prove that }BO=CO\text{ and the ray }OA\text{ bisects}
\displaystyle \angle BAC.
\displaystyle \text{Answer:} 2018-08-11_8-39-15
\displaystyle \text{Since }AB=AC,
\displaystyle \angle ABC=\angle ACB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}

\displaystyle \text{(i) Since }BO\text{ bisects }\angle ABC,
\displaystyle \angle OBC=\frac12\angle ABC.
\displaystyle \text{Since }CO\text{ bisects }\angle ACB,
\displaystyle \angle OCB=\frac12\angle ACB.
\displaystyle \therefore \angle OBC=\angle OCB.
\displaystyle \text{In }\triangle BOC,\text{ sides opposite equal angles are equal.}
\displaystyle \therefore BO=CO.

\displaystyle \text{(ii) Since }BO\text{ and }CO\text{ bisect }\angle ABC\text{ and }\angle ACB,\text{ respectively,}
\displaystyle \angle ABO=\frac12\angle ABC\quad\text{and}\quad\angle ACO=\frac12\angle ACB.
\displaystyle \therefore \angle ABO=\angle ACO.
\displaystyle \text{In }\triangle AOB\text{ and }\triangle AOC,
\displaystyle AB=AC\qquad\text{(Given)}
\displaystyle BO=CO\qquad\text{[Proved in part (i)]}
\displaystyle \angle ABO=\angle ACO
\displaystyle \therefore \triangle AOB\cong\triangle AOC\qquad\text{(SAS congruence criterion)}
\displaystyle \therefore \angle BAO=\angle CAO
\displaystyle \text{(Corresponding parts of congruent triangles are equal.)}
\displaystyle \therefore \text{The ray }OA\text{ bisects }\angle BAC.
\displaystyle \therefore \text{Both the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{In the adjoining figure, it is given that }AB=AC\text{ and }BD=EC.
\displaystyle \text{Prove that }\triangle ABE\cong\triangle ACD. 2018-08-11_8-39-28\displaystyle \text{Answer:}
\displaystyle \text{Since }AB=AC,
\displaystyle \angle ABC=\angle ACB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}
\displaystyle \text{Since }B,D,E,C\text{ are collinear,}
\displaystyle \angle ABE=\angle ABC\quad\text{and}\quad\angle ACD=\angle ACB.
\displaystyle \therefore \angle ABE=\angle ACD.
\displaystyle \text{Also, }BD=EC\qquad\text{(Given)}
\displaystyle \text{Adding }DE\text{ to both sides,}
\displaystyle BD+DE=EC+DE
\displaystyle \therefore BE=CD.
\displaystyle \text{In }\triangle ABE\text{ and }\triangle ACD,
\displaystyle AB=AC\qquad\text{(Given)}
\displaystyle \angle ABE=\angle ACD
\displaystyle BE=CD
\displaystyle \therefore \triangle ABE\cong\triangle ACD\qquad\text{(SAS congruence criterion)}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{In the adjoining figure, the line }l\text{ bisects }\angle QAP\text{ and }B\text{ is any}
\displaystyle \text{point on }l.\ BP\text{ and }BQ\text{ are perpendiculars drawn from }B\text{ to the arms of} 2018-08-11_8-39-48\displaystyle \angle QAP.\text{ Show that:}
\displaystyle \text{(i) }\triangle APB\cong\triangle AQB
\displaystyle \text{(ii) }BP=BQ,\text{ or }B\text{ is equidistant from the arms of }\angle QAP.
\displaystyle \text{Answer:}
\displaystyle \text{Since }B\text{ lies on the bisector }l\text{ of }\angle QAP,
\displaystyle \angle PAB=\angle QAB.
\displaystyle \text{Also, }BP\perp AP\text{ and }BQ\perp AQ.
\displaystyle \therefore \angle APB=\angle AQB=90^\circ.
\displaystyle \text{In }\triangle APB\text{ and }\triangle AQB,
\displaystyle \angle PAB=\angle QAB
\displaystyle \angle APB=\angle AQB
\displaystyle AB=AB\qquad\text{(Common)}
\displaystyle \therefore \triangle APB\cong\triangle AQB\qquad\text{(AAS congruence criterion)}

\displaystyle \text{(ii) Therefore, }BP=BQ
\displaystyle \text{(Corresponding parts of congruent triangles are equal.)}
\displaystyle \text{Since }BP\text{ and }BQ\text{ are the perpendicular distances of }B\text{ from the two arms,}
\displaystyle \therefore B\text{ is equidistant from the arms of }\angle QAP.
\displaystyle \therefore \text{Both the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{In the adjoining figure, }AD\text{ is a median and }BL,\ CM\text{ are perpendiculars}
\displaystyle \text{drawn from }B\text{ and }C\text{ to }AD\text{ and }AD\text{ produced, respectively. Prove that} 2018-08-11_8-40-02\displaystyle BL=CM.
\displaystyle \text{Answer:}
\displaystyle \text{Since }AD\text{ is a median of }\triangle ABC,
\displaystyle BD=DC.
\displaystyle \text{Also, }BL\perp AD\text{ and }CM\perp AD.
\displaystyle \therefore \angle BLD=\angle CMD=90^\circ.
\displaystyle \text{Since the lines }BC\text{ and }LM\text{ intersect at }D,
\displaystyle \angle BDL=\angle CDM\qquad\text{(Vertically opposite angles)}
\displaystyle \text{In }\triangle BDL\text{ and }\triangle CDM,
\displaystyle \angle BLD=\angle CMD
\displaystyle \angle BDL=\angle CDM
\displaystyle BD=DC
\displaystyle \therefore \triangle BDL\cong\triangle CDM\qquad\text{(AAS congruence criterion)}
\displaystyle \therefore BL=CM
\displaystyle \text{(Corresponding parts of congruent triangles are equal.)}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{In the adjoining figure, }BM\perp AC\text{ and }DN\perp AC.\text{ Also,}
\displaystyle BM=DN.\text{ Prove that }AC\text{ bisects }BD. 2018-08-11_8-40-19\displaystyle \text{Answer:}
\displaystyle \text{Let }AC\text{ intersect }BD\text{ at }R.
\displaystyle \text{Since }BM\perp AC\text{ and }DN\perp AC,
\displaystyle \angle BMR=\angle DNR=90^\circ.
\displaystyle \text{Since the lines }AC\text{ and }BD\text{ intersect at }R,
\displaystyle \angle BRM=\angle DRN\qquad\text{(Vertically opposite angles)}
\displaystyle \text{In }\triangle BMR\text{ and }\triangle DNR,
\displaystyle \angle BMR=\angle DNR
\displaystyle \angle BRM=\angle DRN
\displaystyle BM=DN\qquad\text{(Given)}
\displaystyle \therefore \triangle BMR\cong\triangle DNR\qquad\text{(AAS congruence criterion)}
\displaystyle \therefore BR=DR
\displaystyle \text{(Corresponding parts of congruent triangles are equal.)}
\displaystyle \therefore R\text{ is the midpoint of }BD.
\displaystyle \text{Since }R\text{ lies on }AC,\ AC\text{ bisects }BD.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{In the adjoining figure, }\triangle ABC\text{ is isosceles with }AB=AC.
\displaystyle BD\text{ and }CE\text{ are two medians of the triangle. Prove that }BD=CE. 2018-08-11_8-40-32\displaystyle \text{Answer:}
\displaystyle \text{Since }BD\text{ and }CE\text{ are medians, }D\text{ and }E\text{ are the midpoints of }AC\text{ and }AB.
\displaystyle \therefore DC=\frac12AC\quad\text{and}\quad BE=\frac12AB.
\displaystyle \text{Since }AB=AC,
\displaystyle \frac12AB=\frac12AC.
\displaystyle \therefore BE=DC.
\displaystyle \text{Also, since }AB=AC,
\displaystyle \angle ABC=\angle ACB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}
\displaystyle \text{Since }E\text{ lies on }AB\text{ and }D\text{ lies on }AC,
\displaystyle \angle EBC=\angle ABC\quad\text{and}\quad\angle DCB=\angle ACB.
\displaystyle \therefore \angle EBC=\angle DCB.
\displaystyle \text{In }\triangle BEC\text{ and }\triangle CDB,
\displaystyle BE=DC
\displaystyle \angle EBC=\angle DCB
\displaystyle BC=CB\qquad\text{(Common)}
\displaystyle \therefore \triangle BEC\cong\triangle CDB\qquad\text{(SAS congruence criterion)}
\displaystyle \therefore CE=BD
\displaystyle \text{(Corresponding parts of congruent triangles are equal.)}
\displaystyle \therefore BD=CE.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 31: }\text{In the adjoining figure, }PS=PR\text{ and }\angle TPS=\angle QPR.
\displaystyle \text{Prove that }PT=PQ. 2018-08-11_8-40-44\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle PSR,
\displaystyle PS=PR\qquad\text{(Given)}
\displaystyle \therefore \angle PSR=\angle PRS
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}
\displaystyle \text{Since }T,S,R\text{ are collinear,}
\displaystyle \angle PST+\angle PSR=180^\circ.
\displaystyle \text{Since }S,R,Q\text{ are collinear,}
\displaystyle \angle PRS+\angle PRQ=180^\circ.
\displaystyle \text{Since }\angle PSR=\angle PRS,
\displaystyle \therefore \angle PST=\angle PRQ.
\displaystyle \text{In }\triangle PST\text{ and }\triangle PRQ,
\displaystyle \angle PST=\angle PRQ
\displaystyle PS=PR\qquad\text{(Given)}
\displaystyle \angle TPS=\angle QPR\qquad\text{(Given)}
\displaystyle \therefore \triangle PST\cong\triangle PRQ\qquad\text{(ASA congruence criterion)}
\displaystyle \therefore PT=PQ
\displaystyle \text{(Corresponding parts of congruent triangles are equal.)}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{In the adjoining figure, }\triangle ABC\text{ and }\triangle DBC\text{ are two triangles}
\displaystyle \text{on the same base }BC\text{ such that }AB=AC\text{ and }DB=DC.\text{ Prove that} 2018-08-11_8-40-59\displaystyle \angle ABD=\angle ACD.
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,
\displaystyle AB=AC\qquad\text{(Given)}
\displaystyle \therefore \angle ABC=\angle ACB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}\qquad\ldots\text{(i)}
\displaystyle \text{In }\triangle DBC,
\displaystyle DB=DC\qquad\text{(Given)}
\displaystyle \therefore \angle DBC=\angle DCB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}\qquad\ldots\text{(ii)}
\displaystyle \text{Since }BD\text{ lies inside }\angle ABC,
\displaystyle \angle ABD=\angle ABC-\angle DBC.
\displaystyle \text{Since }CD\text{ lies inside }\angle ACB,
\displaystyle \angle ACD=\angle ACB-\angle DCB.
\displaystyle \text{Subtracting the equal angles in (ii) from the equal angles in (i), we get}
\displaystyle \angle ABC-\angle DBC=\angle ACB-\angle DCB
\displaystyle \therefore \angle ABD=\angle ACD.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{In the adjoining figure, }AB\parallel CD\text{ and }O\text{ is the midpoint of }AD.
\displaystyle \text{Show that:}
\displaystyle \text{(i) }\triangle AOB\cong\triangle DOC\qquad\text{(ii) }O\text{ is also the midpoint of }BC. 2018-08-11_8-41-07\displaystyle \text{Answer:}
\displaystyle \text{(i) Since }O\text{ is the midpoint of }AD,
\displaystyle OA=OD.
\displaystyle \text{Since }AB\parallel CD\text{ and }BC\text{ is a transversal,}
\displaystyle \angle ABO=\angle DCO\qquad\text{(Alternate interior angles)}
\displaystyle \text{Also, since the lines }AD\text{ and }BC\text{ intersect at }O,
\displaystyle \angle AOB=\angle DOC\qquad\text{(Vertically opposite angles)}
\displaystyle \text{In }\triangle AOB\text{ and }\triangle DOC,
\displaystyle OA=OD
\displaystyle \angle ABO=\angle DCO
\displaystyle \angle AOB=\angle DOC
\displaystyle \therefore \triangle AOB\cong\triangle DOC\qquad\text{(AAS congruence criterion)}

\displaystyle \text{(ii) From }\triangle AOB\cong\triangle DOC,
\displaystyle OB=OC
\displaystyle \text{(Corresponding parts of congruent triangles are equal.)}
\displaystyle \text{Since }B,\ O\text{ and }C\text{ are collinear and }OB=OC,
\displaystyle \therefore O\text{ is the midpoint of }BC.
\displaystyle \therefore \text{Both the required results are proved.}
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{In }\triangle ABC,\text{ it is given that }AB=AC\text{ and the bisectors of}
\displaystyle \angle B\text{ and }\angle C\text{ intersect at }O.\text{ If }M\text{ is a point on }BO\text{ produced, prove}
\displaystyle \text{that }\angle MOC=\angle ABC.
\displaystyle \text{Answer:} 2018-08-11_8-48-31.jpg
\displaystyle \text{Since }AB=AC,
\displaystyle \angle ABC=\angle ACB
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}\qquad\ldots\text{(i)}
\displaystyle \text{Since }BO\text{ bisects }\angle ABC,
\displaystyle \angle OBC=\frac12\angle ABC.
\displaystyle \text{Since }CO\text{ bisects }\angle ACB,
\displaystyle \angle OCB=\frac12\angle ACB.
\displaystyle \therefore \angle OBC=\angle OCB\qquad\text{[Using (i)]}
\displaystyle \text{Since }B,\ O\text{ and }M\text{ are collinear, }\angle MOC\text{ is an exterior angle of}
\displaystyle \triangle OBC.
\displaystyle \therefore \angle MOC=\angle OBC+\angle OCB
\displaystyle =2\angle OBC
\displaystyle =2\left(\frac12\angle ABC\right)
\displaystyle \therefore \angle MOC=\angle ABC.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 35: }P\text{ is a point on the bisector of }\angle ABC.\text{ The line through }P
\displaystyle \text{parallel to }AB\text{ meets }BC\text{ at }Q.\text{ Prove that }\triangle BPQ\text{ is isosceles.}
\displaystyle \text{Answer:}
\displaystyle \text{Since }PQ\parallel AB\text{ and }BP\text{ is a transversal,}
\displaystyle \angle ABP=\angle BPQ\qquad\text{(Alternate interior angles)}
\displaystyle \text{Since }BP\text{ bisects }\angle ABC,2018-08-11_8-48-48
\displaystyle \angle ABP=\angle PBC.
\displaystyle \text{Since }Q\text{ lies on }BC,
\displaystyle \angle PBC=\angle PBQ.
\displaystyle \therefore \angle BPQ=\angle PBQ.
\displaystyle \text{In }\triangle BPQ,\text{ sides opposite equal angles are equal.}
\displaystyle \therefore BQ=PQ.
\displaystyle \therefore \triangle BPQ\text{ is an isosceles triangle.}
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

\displaystyle \textbf{Question 36: }\triangle ABC\text{ is a triangle in which }\angle B=2\angle C.\ D\text{ is a point on}
\displaystyle BC\text{ such that }AD\text{ bisects }\angle BAC\text{ and }AB=CD.\text{ Prove that}
\displaystyle \angle BAC=72^\circ.
\displaystyle \text{Answer:} 2018-08-11_8-56-41
\displaystyle \text{Let }\angle ACB=y.
\displaystyle \therefore \angle ABC=2y.
\displaystyle \text{Since }AD\text{ bisects }\angle BAC,\text{ let}
\displaystyle \angle BAD=\angle DAC=x.
\displaystyle \therefore \angle BAC=2x.
\displaystyle \text{Let the bisector of }\angle ABC\text{ meet }AC\text{ at }P.
\displaystyle \therefore \angle ABP=\angle PBC=y.
\displaystyle \text{Since }P\text{ lies on }AC,
\displaystyle \angle BCP=\angle BCA=y.
\displaystyle \therefore \angle PBC=\angle BCP.
\displaystyle \therefore BP=PC
\displaystyle \text{(Sides opposite equal angles of a triangle are equal.)}
\displaystyle \text{In }\triangle ABP\text{ and }\triangle DCP,
\displaystyle AB=CD\qquad\text{(Given)}
\displaystyle BP=CP
\displaystyle \angle ABP=\angle DCP=y
\displaystyle \therefore \triangle ABP\cong\triangle DCP\qquad\text{(SAS congruence criterion)}
\displaystyle \therefore AP=DP\quad\text{and}\quad\angle BAP=\angle CDP
\displaystyle \text{(Corresponding parts of congruent triangles are equal.)}
\displaystyle \text{Since }P\text{ lies on }AC,
\displaystyle \angle BAP=\angle BAC=2x.
\displaystyle \therefore \angle CDP=2x.
\displaystyle \text{Also, }AP=DP.
\displaystyle \therefore \angle PAD=\angle ADP
\displaystyle \text{(Angles opposite equal sides of a triangle are equal.)}
\displaystyle \text{Since }P\text{ lies on }AC,\ \angle PAD=\angle CAD=x.
\displaystyle \therefore \angle ADP=x.
\displaystyle \text{Since }B,D,C\text{ are collinear,}
\displaystyle \angle BDA+\angle ADP+\angle PDC=180^\circ.
\displaystyle \therefore \angle BDA+x+2x=180^\circ
\displaystyle \therefore \angle BDA=180^\circ-3x.
\displaystyle \text{In }\triangle ABD,
\displaystyle \angle BAD+\angle ABD+\angle BDA=180^\circ
\displaystyle x+2y+(180^\circ-3x)=180^\circ
\displaystyle \therefore 2y=2x
\displaystyle \therefore x=y.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle BAC+\angle ABC+\angle ACB=180^\circ
\displaystyle 2x+2y+y=180^\circ
\displaystyle \text{Since }x=y,
\displaystyle 5x=180^\circ
\displaystyle \therefore x=36^\circ.
\displaystyle \therefore \angle BAC=2x=72^\circ.
\displaystyle \therefore \text{The required result is proved.}
\displaystyle \\

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