\displaystyle \textbf{Question 1: }\text{From the following figure, prove that }AB>CD. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }AB=AC,\text{ the angles opposite to these equal sides are equal.}
\displaystyle \therefore \angle ABC=\angle ACB=70^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle BAC=180^\circ-(70^\circ+70^\circ)=40^\circ.
\displaystyle \text{Since }B,\ C\text{ and }D\text{ are collinear, }\angle ABD=70^\circ.
\displaystyle \text{In }\triangle ABD,
\displaystyle \angle BAD=180^\circ-(70^\circ+40^\circ)=70^\circ.
\displaystyle \therefore \angle CAD=\angle BAD-\angle BAC=70^\circ-40^\circ=30^\circ.
\displaystyle \text{In }\triangle ACD,\quad \angle ADC=40^\circ>\angle CAD=30^\circ.
\displaystyle \text{The side opposite the greater angle is greater.}
\displaystyle \therefore AC>CD.
\displaystyle \text{But }AB=AC.
\displaystyle {\therefore AB>CD.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In }\triangle PQR,\ QR=PR\text{ and }\angle P=36^\circ.
\displaystyle \text{Which is the largest side of the triangle?}
\displaystyle \textbf{Answer:} \displaystyle \text{Since }QR=PR,\text{ the angles opposite these equal sides are equal.}
\displaystyle \therefore \angle Q=\angle P=36^\circ.
\displaystyle \text{In }\triangle PQR,
\displaystyle \angle R=180^\circ-(36^\circ+36^\circ)=108^\circ.
\displaystyle \therefore \angle R>\angle P\text{ and }\angle R>\angle Q.
\displaystyle \text{The side opposite the greatest angle is the greatest side.}
\displaystyle {\therefore PQ\text{ is the largest side of }\triangle PQR.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If two sides of a triangle are }8\text{ cm and }13\text{ cm,}
\displaystyle \text{then the length of the third side is between }a\text{ cm and }b\text{ cm.}
\displaystyle \text{Find }a\text{ and }b,\text{ such that }a<b.
\displaystyle \textbf{Answer:}
\displaystyle \text{Let the length of the third side be }x\text{ cm.}
\displaystyle \text{The sum of any two sides of a triangle is greater than the third side.}
\displaystyle \therefore 8+13>x.
\displaystyle \therefore x<21.
\displaystyle \text{Also, the difference of any two sides is less than the third side.}
\displaystyle \therefore 13-8<x.
\displaystyle \therefore 5<x.
\displaystyle \therefore 5<x<21.
\displaystyle \text{Comparing with }a<x<b,\text{ we get }a=5\text{ and }b=21.
\displaystyle {\therefore a=5\text{ and }b=21.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In each of the following figures, write }BC,\ AC
\displaystyle \text{and }CD\text{ in ascending order of their lengths.} \displaystyle \textbf{Answer:}
\displaystyle \text{(i) Since }AB=AC,\text{ the angles opposite these equal sides are equal.}
\displaystyle \therefore \angle ABC=\angle ACB=67^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle BAC=180^\circ-(67^\circ+67^\circ)=46^\circ.
\displaystyle \text{Since }\angle ACB=67^\circ>\angle BAC=46^\circ,
\displaystyle \therefore AB>BC.
\displaystyle \text{But }AB=AC.
\displaystyle \therefore AC>BC.\qquad\ldots\text{(I)}
\displaystyle \text{Since }B,\ C\text{ and }D\text{ are collinear,}
\displaystyle \angle ACD=180^\circ-\angle ACB=180^\circ-67^\circ=113^\circ.
\displaystyle \text{In }\triangle ACD,
\displaystyle \angle CAD=180^\circ-(113^\circ+33^\circ)=34^\circ.
\displaystyle \text{Since }\angle CAD=34^\circ>\angle ADC=33^\circ,
\displaystyle \therefore CD>AC.\qquad\ldots\text{(II)}
\displaystyle \text{From (I) and (II),}
\displaystyle {\therefore BC<AC<CD.}
\displaystyle \text{Hence, the ascending order is }BC,\ AC,\ CD.
\displaystyle \\

\displaystyle \text{(ii) In }\triangle ABC,
\displaystyle \angle ACB=180^\circ-(47^\circ+73^\circ)=60^\circ.
\displaystyle \text{Since }\angle ABC=73^\circ>\angle BAC=47^\circ,
\displaystyle \therefore AC>BC.\qquad\ldots\text{(I)}
\displaystyle \text{Since }B,\ C\text{ and }D\text{ are collinear,}
\displaystyle \angle ACD=180^\circ-\angle ACB=180^\circ-60^\circ=120^\circ.
\displaystyle \text{In }\triangle ACD,
\displaystyle \angle ADC=180^\circ-(120^\circ+31^\circ)=29^\circ.
\displaystyle \text{Since }\angle CAD=31^\circ>\angle ADC=29^\circ,
\displaystyle \therefore CD>AC.\qquad\ldots\text{(II)}
\displaystyle \text{From (I) and (II),}
\displaystyle {\therefore BC<AC<CD.}
\displaystyle \text{Hence, the ascending order is }BC,\ AC,\ CD.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Arrange the sides of }\triangle BOC\text{ in descending}
\displaystyle \text{order of their lengths. }BO\text{ and }CO\text{ are the bisectors of}
\displaystyle \angle ABC\text{ and }\angle ACB\text{ respectively.} \displaystyle \textbf{Answer:}
\displaystyle \text{Since }D,\ A\text{ and }C\text{ are collinear,}
\displaystyle \angle BAC=180^\circ-137^\circ=43^\circ.
\displaystyle \text{Since }E,\ B\text{ and }C\text{ are collinear,}
\displaystyle \angle ABC=180^\circ-106^\circ=74^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle ACB=180^\circ-(43^\circ+74^\circ)=63^\circ.
\displaystyle \text{Since }BO\text{ bisects }\angle ABC,
\displaystyle \angle OBC=\frac{1}{2}\angle ABC=\frac{74^\circ}{2}=37^\circ.
\displaystyle \text{Since }CO\text{ bisects }\angle ACB,
\displaystyle \angle BCO=\frac{1}{2}\angle ACB=\frac{63^\circ}{2}=31.5^\circ.
\displaystyle \text{In }\triangle BOC,
\displaystyle \angle BOC=180^\circ-(37^\circ+31.5^\circ)=111.5^\circ.
\displaystyle \therefore \angle BOC>\angle OBC>\angle BCO.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle {\therefore BC>CO>BO.}
\displaystyle \text{Hence, the descending order is }BC,\ CO,\ BO.
\displaystyle \\

\displaystyle \textbf{Question 6: }D\text{ is a point on side }BC\text{ of }\triangle ABC.
\displaystyle \text{If }AD>AC,\text{ show that }AB>AC.
\displaystyle \textbf{Answer:} \displaystyle \text{In }\triangle ACD,\quad AD>AC.
\displaystyle \therefore \angle ACD>\angle ADC.
\displaystyle \text{Since }B,\ D\text{ and }C\text{ are collinear, }\angle ADC\text{ is an exterior angle}
\displaystyle \text{of }\triangle ABD.
\displaystyle \therefore \angle ADC>\angle ABD.
\displaystyle \therefore \angle ACD>\angle ABD.
\displaystyle \text{Since }D\text{ lies on }BC,
\displaystyle \angle ACD=\angle ACB\text{ and }\angle ABD=\angle ABC.
\displaystyle \therefore \angle ACB>\angle ABC.
\displaystyle \text{In }\triangle ABC,\text{ the greater angle has the greater side opposite to it.}
\displaystyle {\therefore AB>AC.}
\displaystyle \\

\displaystyle \textbf{Question 7: }O\text{ is a point in the interior of }\triangle ABC.
\displaystyle \text{Show that }OB+OC<AB+AC.
\displaystyle \textbf{Answer:} \displaystyle \text{Produce }CO\text{ to meet }AB\text{ at }M.
\displaystyle \text{In }\triangle ACM,\text{ the sum of any two sides is greater than the third side.}
\displaystyle AC+AM>CM.
\displaystyle \therefore AC+AM>CO+OM.\qquad\ldots\text{(I)}
\displaystyle \text{In }\triangle BOM,\text{ the sum of any two sides is greater than the third side.}
\displaystyle BM+OM>BO.\qquad\ldots\text{(II)}
\displaystyle \text{Adding (I) and (II), we get}
\displaystyle AC+AM+BM+OM>CO+OM+BO.
\displaystyle \therefore AC+(AM+BM)>CO+BO.
\displaystyle \text{Since }AM+BM=AB,
\displaystyle AC+AB>OC+OB.
\displaystyle {\therefore OB+OC<AB+AC.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In the following figure, }\angle BAC=60^\circ\text{ and}
\displaystyle \angle ABC=65^\circ.\text{ Prove that:}
\displaystyle \text{(i) }CF>AF\qquad\text{(ii) }DC>DF. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }AD\perp BC,\quad \angle ADB=90^\circ.
\displaystyle \text{In }\triangle ABD,
\displaystyle \angle BAD=180^\circ-(90^\circ+65^\circ)=25^\circ.
\displaystyle \therefore \angle FAC=\angle BAC-\angle BAF=60^\circ-25^\circ=35^\circ.
\displaystyle \text{Since }CE\perp AB,\quad \angle AEC=90^\circ.
\displaystyle \text{In }\triangle AEC,
\displaystyle \angle ACE=180^\circ-(90^\circ+60^\circ)=30^\circ.
\displaystyle \text{Since }F\text{ lies on }CE,\quad \angle ACF=\angle ACE=30^\circ.
\displaystyle \text{(i) In }\triangle ACF,\quad \angle FAC=35^\circ>\angle ACF=30^\circ.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle {\therefore CF>AF.}
\displaystyle \text{(ii) In }\triangle ABC,
\displaystyle \angle ACB=180^\circ-(60^\circ+65^\circ)=55^\circ.
\displaystyle \therefore \angle DCF=\angle BCE=\angle BCA-\angle ECA.
\displaystyle \therefore \angle DCF=55^\circ-30^\circ=25^\circ.
\displaystyle \text{Since }AD\perp BC,\quad \angle FDC=90^\circ.
\displaystyle \text{In }\triangle DFC,
\displaystyle \angle DFC=180^\circ-(90^\circ+25^\circ)=65^\circ.
\displaystyle \therefore \angle DFC=65^\circ>\angle DCF=25^\circ.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle {\therefore DC>DF.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{In the following figure, }AC=CD,\ \angle BAD=110^\circ
\displaystyle \text{and }\angle ACB=74^\circ.\text{ Prove that }BC>CD. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }B,\ C\text{ and }D\text{ are collinear,}
\displaystyle \angle ACD=180^\circ-\angle ACB=180^\circ-74^\circ=106^\circ.
\displaystyle \text{In }\triangle ACD,\quad AC=CD.
\displaystyle \therefore \angle CAD=\angle ADC.
\displaystyle \therefore \angle CAD=\angle ADC=\frac{180^\circ-106^\circ}{2}=37^\circ.
\displaystyle \angle BAC=\angle BAD-\angle CAD=110^\circ-37^\circ=73^\circ.
\displaystyle \text{In }\triangle ABC,
\displaystyle \angle ABC=180^\circ-(73^\circ+74^\circ)=33^\circ.
\displaystyle \therefore \angle BAC=73^\circ>\angle ABC=33^\circ.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle \therefore BC>AC.
\displaystyle \text{But }AC=CD.
\displaystyle {\therefore BC>CD.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{From the following figure, prove that:}
\displaystyle \text{(i) }AB>BD\qquad\text{(ii) }AC>CD\qquad\text{(iii) }AB+AC>BC. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }AD\perp BC,\quad \angle ADB=\angle ADC=90^\circ.
\displaystyle \text{(i) In right-angled }\triangle ABD,\quad \angle ADB>\angle BAD.
\displaystyle \text{The side opposite the greater angle is greater.}
\displaystyle {\therefore AB>BD.}
\displaystyle \text{(ii) In right-angled }\triangle ACD,\quad \angle ADC>\angle CAD.
\displaystyle \text{The side opposite the greater angle is greater.}
\displaystyle {\therefore AC>CD.}
\displaystyle \text{(iii) Adding }AB>BD\text{ and }AC>CD,\text{ we get}
\displaystyle AB+AC>BD+CD.
\displaystyle \text{Since }D\text{ lies on }BC,\quad BD+CD=BC.
\displaystyle {\therefore AB+AC>BC.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In a quadrilateral }ABCD,\text{ prove that:}
\displaystyle \text{(i) }AB+BC+CD>DA
\displaystyle \text{(ii) }AB+BC+CD+DA>2AC
\displaystyle \text{(iii) }AB+BC+CD+DA>2BD
\displaystyle \textbf{Answer:} \displaystyle \text{Join }A\text{ and }C.
\displaystyle \text{(i) In }\triangle ABC,
\displaystyle AB+BC>AC.\qquad\ldots\text{(I)}
\displaystyle \text{In }\triangle ACD,
\displaystyle AC+CD>AD.\qquad\ldots\text{(II)}
\displaystyle \text{Adding (I) and (II), we get}
\displaystyle AB+BC+AC+CD>AC+AD.
\displaystyle \therefore AB+BC+CD>AD.
\displaystyle {\therefore AB+BC+CD>DA.}
\displaystyle \text{(ii) In }\triangle ABC,
\displaystyle AB+BC>AC.\qquad\ldots\text{(III)}
\displaystyle \text{In }\triangle ACD,
\displaystyle AD+CD>AC.\qquad\ldots\text{(IV)}
\displaystyle \text{Adding (III) and (IV), we get}
\displaystyle AB+BC+CD+DA>2AC.
\displaystyle {\therefore AB+BC+CD+DA>2AC.}
\displaystyle \text{(iii) Join }B\text{ and }D.
\displaystyle \text{In }\triangle ABD,
\displaystyle AB+AD>BD.\qquad\ldots\text{(V)}
\displaystyle \text{In }\triangle BCD,
\displaystyle BC+CD>BD.\qquad\ldots\text{(VI)}
\displaystyle \text{Adding (V) and (VI), we get}
\displaystyle AB+BC+CD+DA>2BD.
\displaystyle {\therefore AB+BC+CD+DA>2BD.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the following figure, }ABC\text{ is an equilateral}
\displaystyle \text{triangle and }P\text{ is any point on }AC.\text{ Prove that:}
\displaystyle \text{(i) }BP>PA\qquad\text{(ii) }BP>PC. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }\triangle ABC\text{ is equilateral,}
\displaystyle \angle BAC=\angle ABC=\angle ACB=60^\circ.
\displaystyle \text{Since }P\text{ lies on }AC,\quad \angle BAP=\angle BAC=60^\circ.
\displaystyle \text{(i) In }\triangle ABP,
\displaystyle \angle APB\text{ is an exterior angle of }\triangle BPC.
\displaystyle \therefore \angle APB>\angle BCP.
\displaystyle \text{But }\angle BCP=\angle BCA=60^\circ.
\displaystyle \therefore \angle APB>60^\circ=\angle BAP.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle {\therefore BP>PA.}
\displaystyle \text{(ii) Since }P\text{ lies on }AC,\quad \angle BCP=\angle BCA=60^\circ.
\displaystyle \text{In }\triangle BPC,\quad \angle BPC\text{ is an exterior angle of }\triangle ABP.
\displaystyle \therefore \angle BPC>\angle BAP.
\displaystyle \therefore \angle BPC>60^\circ=\angle BCP.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle {\therefore BC>BP.}
\displaystyle \text{This does not prove }BP>PC,\text{ so we compare the correct angles in }\triangle BPC.
\displaystyle \angle PBC=\angle ABC-\angle ABP.
\displaystyle \text{Also, in }\triangle ABP,\quad \angle APB=120^\circ+\angle ABP.
\displaystyle \text{Since }\angle BPC=180^\circ-\angle APB,
\displaystyle \angle BPC=60^\circ-\angle ABP=\angle PBC.
\displaystyle \therefore BP=PC.
\displaystyle \text{Hence, the statement }BP>PC\text{ is not true for every point }P\text{ on }AC.
\displaystyle \\

\displaystyle \textbf{Question 13: }P\text{ is any point inside }\triangle ABC.\text{ Prove that}
\displaystyle \angle BPC>\angle BAC.
\displaystyle \textbf{Answer:} \displaystyle \text{Produce }BP\text{ to meet }AC\text{ at }D.
\displaystyle \text{In }\triangle PCD,\quad \angle BPC\text{ is an exterior angle.}
\displaystyle \therefore \angle BPC>\angle PDC.
\displaystyle \text{Since }B,\ P\text{ and }D\text{ are collinear,}
\displaystyle \angle PDC=\angle BDC.
\displaystyle \text{In }\triangle ABD,\quad \angle BDC\text{ is an exterior angle.}
\displaystyle \therefore \angle BDC>\angle BAD.
\displaystyle \text{Since }D\text{ lies on }AC,\quad \angle BAD=\angle BAC.
\displaystyle \therefore \angle BPC>\angle BDC>\angle BAC.
\displaystyle {\therefore \angle BPC>\angle BAC.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Prove that the straight line joining the vertex of an}
\displaystyle \text{isosceles triangle to any point on its base is smaller than either}
\displaystyle \text{of the equal sides of the triangle.}
\displaystyle \textbf{Answer:} \displaystyle \text{Let }AB=AC\text{ in isosceles }\triangle ABC,\text{ and let }P\text{ be any point on }BC.
\displaystyle \text{Since }AB=AC,\text{ the angles opposite the equal sides are equal.}
\displaystyle \therefore \angle ABC=\angle ACB.
\displaystyle \text{Since }P\text{ lies on }BC,
\displaystyle \angle ABP=\angle ABC\text{ and }\angle ACP=\angle ACB.
\displaystyle \therefore \angle ABP=\angle ACP.
\displaystyle \text{In }\triangle APC,\quad \angle APB\text{ is an exterior angle.}
\displaystyle \therefore \angle APB>\angle ACP.
\displaystyle \therefore \angle APB>\angle ABP.
\displaystyle \text{In }\triangle ABP,\text{ the greater angle has the greater side opposite to it.}
\displaystyle \therefore AB>AP.\qquad\ldots\text{(I)}
\displaystyle \text{In }\triangle ABP,\quad \angle APC\text{ is an exterior angle.}
\displaystyle \therefore \angle APC>\angle ABP.
\displaystyle \therefore \angle APC>\angle ACP.
\displaystyle \text{In }\triangle APC,\text{ the greater angle has the greater side opposite to it.}
\displaystyle \therefore AC>AP.\qquad\ldots\text{(II)}
\displaystyle \text{From (I) and (II), }AP<AB\text{ and }AP<AC.
\displaystyle {\therefore AP\text{ is smaller than either of the equal sides.}}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In the following diagram, }AD=AB\text{ and }AE
\displaystyle \text{bisects }\angle A.\text{ Prove that:}
\displaystyle \text{(i) }BE=DE\qquad\text{(ii) }\angle ABD>\angle C. \displaystyle \textbf{Answer:}
\displaystyle \text{Since }AE\text{ bisects }\angle BAC,
\displaystyle \angle BAE=\angle EAD.
\displaystyle \text{(i) In }\triangle ABE\text{ and }\triangle ADE,
\displaystyle AB=AD\qquad\text{(Given)}
\displaystyle AE=AE\qquad\text{(Common)}
\displaystyle \angle BAE=\angle EAD\qquad\text{(Since }AE\text{ bisects }\angle BAC\text{)}
\displaystyle \therefore \triangle ABE\cong\triangle ADE\qquad\text{(By S.A.S.)}
\displaystyle {\therefore BE=DE.}\qquad\text{(By C.P.C.T.C.)}
\displaystyle \text{(ii) Since }AB=AD,\text{ the angles opposite these equal sides are equal.}
\displaystyle \therefore \angle ABD=\angle BDA.
\displaystyle \text{In }\triangle BDC,\quad \angle BDA\text{ is an exterior angle.}
\displaystyle \therefore \angle BDA>\angle BCD.
\displaystyle \text{Since }D\text{ lies on }AC,\quad \angle BCD=\angle BCA=\angle C.
\displaystyle \therefore \angle BDA>\angle C.
\displaystyle \text{But }\angle ABD=\angle BDA.
\displaystyle {\therefore \angle ABD>\angle C.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The sides }AB\text{ and }AC\text{ of }\triangle ABC\text{ are}
\displaystyle \text{produced, and the bisectors of the exterior angles at }B\text{ and }C
\displaystyle \text{meet at }P.\text{ Prove that if }AB>AC,\text{ then }PC>PB.
\displaystyle \textbf{Answer:} \displaystyle \text{Since }AB>AC,\text{ the angle opposite }AB\text{ is greater than the angle}
\displaystyle \text{opposite }AC.
\displaystyle \therefore \angle ACB>\angle ABC.
\displaystyle \text{Let the exterior angles at }B\text{ and }C\text{ be }\angle ABP\text{ and }\angle ACP.
\displaystyle \text{Since }BP\text{ bisects the exterior angle at }B,
\displaystyle \angle PBC=\frac{1}{2}\left(180^\circ-\angle ABC\right).
\displaystyle \text{Since }CP\text{ bisects the exterior angle at }C,
\displaystyle \angle BCP=\frac{1}{2}\left(180^\circ-\angle ACB\right).
\displaystyle \text{But }\angle ACB>\angle ABC.
\displaystyle \therefore 180^\circ-\angle ABC>180^\circ-\angle ACB.
\displaystyle \therefore \angle PBC>\angle BCP.
\displaystyle \text{In }\triangle PBC,\text{ the greater angle has the greater side opposite to it.}
\displaystyle {\therefore PC>PB.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In the following figure, }AB\text{ is the largest side and}
\displaystyle BC\text{ is the smallest side of }\triangle ABC.\text{ Write the angles }x^\circ,\ y^\circ
\displaystyle \text{and }z^\circ\text{ in ascending order of their values.} \displaystyle \textbf{Answer:}
\displaystyle \text{Since }AB\text{ is the largest side, }\angle C\text{ is the greatest angle.}
\displaystyle \text{Since }BC\text{ is the smallest side, }\angle A\text{ is the smallest angle.}
\displaystyle \therefore \angle A<\angle B<\angle C.
\displaystyle \text{Since each exterior angle and its adjacent interior angle are supplementary,}
\displaystyle x^\circ=180^\circ-\angle A,
\displaystyle y^\circ=180^\circ-\angle B,
\displaystyle z^\circ=180^\circ-\angle C.
\displaystyle \text{As }\angle A<\angle B<\angle C,\text{ their supplements are in the reverse order.}
\displaystyle \therefore 180^\circ-\angle C<180^\circ-\angle B<180^\circ-\angle A.
\displaystyle {\therefore z^\circ<y^\circ<x^\circ.}
\displaystyle \text{Hence, the ascending order is }z^\circ,\ y^\circ,\ x^\circ.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In quadrilateral }ABCD,\text{ side }AB\text{ is the longest}
\displaystyle \text{and side }DC\text{ is the shortest. Prove that:}
\displaystyle \text{(i) }\angle C>\angle A\qquad\text{(ii) }\angle D>\angle B.
\displaystyle \textbf{Answer:} \displaystyle \text{(i) Join }A\text{ and }C.
\displaystyle \text{In }\triangle ABC,\quad AB>BC\qquad\text{(Since }AB\text{ is the longest side)}
\displaystyle \therefore \angle ACB>\angle BAC.\qquad\ldots\text{(I)}
\displaystyle \text{In }\triangle ACD,\quad AD>CD\qquad\text{(Since }CD\text{ is the shortest side)}
\displaystyle \therefore \angle ACD>\angle CAD.\qquad\ldots\text{(II)}
\displaystyle \text{Adding (I) and (II), we get}
\displaystyle \angle ACB+\angle ACD>\angle BAC+\angle CAD.
\displaystyle \therefore \angle BCD>\angle BAD.
\displaystyle {\therefore \angle C>\angle A.}
\displaystyle \text{(ii) Join }B\text{ and }D.
\displaystyle \text{In }\triangle ABD,\quad AB>AD\qquad\text{(Since }AB\text{ is the longest side)}
\displaystyle \therefore \angle ADB>\angle ABD.\qquad\ldots\text{(III)}
\displaystyle \text{In }\triangle BCD,\quad BC>CD\qquad\text{(Since }CD\text{ is the shortest side)}
\displaystyle \therefore \angle BDC>\angle DBC.\qquad\ldots\text{(IV)}
\displaystyle \text{Adding (III) and (IV), we get}
\displaystyle \angle ADB+\angle BDC>\angle ABD+\angle DBC.
\displaystyle \therefore \angle ADC>\angle ABC.
\displaystyle {\therefore \angle D>\angle B.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The following figure shows }\triangle ABC\text{ with exterior}
\displaystyle \text{angles }x,\ y\text{ and }z.
\displaystyle \text{(i) If }AB>AC>BC,\text{ arrange }x,\ y\text{ and }z\text{ in ascending order.}
\displaystyle \text{(ii) If }y>x>z,\text{ arrange }AB,\ BC\text{ and }AC\text{ in descending order.} \displaystyle \textbf{Answer:}
\displaystyle \text{(i) Since }AB>AC>BC,\text{ the angles opposite these sides satisfy}
\displaystyle \angle C>\angle B>\angle A.
\displaystyle \text{Each exterior angle and its adjacent interior angle are supplementary.}
\displaystyle x=180^\circ-\angle A,\qquad y=180^\circ-\angle B,
\displaystyle z=180^\circ-\angle C.
\displaystyle \text{The supplements of greater angles are smaller.}
\displaystyle \therefore 180^\circ-\angle C<180^\circ-\angle B<180^\circ-\angle A.
\displaystyle {\therefore z<y<x.}
\displaystyle \text{Hence, the ascending order is }z,\ y,\ x.
\displaystyle \text{(ii) Given }y>x>z.
\displaystyle \therefore 180^\circ-\angle B>180^\circ-\angle A>180^\circ-\angle C.
\displaystyle \therefore \angle B<\angle A<\angle C.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle \therefore AB>BC>AC.
\displaystyle {\therefore \text{The descending order is }AB,\ BC,\ AC.}
\displaystyle \\

\displaystyle \textbf{Question 20:}
\displaystyle \text{(i) In a right-angled triangle, prove that the hypotenuse is the greatest side.}
\displaystyle \text{(ii) In }\triangle ABC,\ \angle ACB=108^\circ.\text{ Show that }AB\text{ is the largest side.}
\displaystyle \textbf{Answer:}
\displaystyle \text{(i) Let }\triangle ABC\text{ be right-angled at }C.\displaystyle \therefore \angle C=90^\circ.
\displaystyle \text{The other two angles of a triangle are acute.}
\displaystyle \therefore \angle C>\angle A\text{ and }\angle C>\angle B.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle \therefore AB>BC\text{ and }AB>AC.
\displaystyle {\therefore AB,\text{ the hypotenuse, is the greatest side.}}
\displaystyle \text{(ii) In }\triangle ABC,\quad \angle ACB=108^\circ.\displaystyle \angle A+\angle B=180^\circ-108^\circ=72^\circ.
\displaystyle \therefore \angle A<108^\circ\text{ and }\angle B<108^\circ.
\displaystyle \therefore \angle C>\angle A\text{ and }\angle C>\angle B.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle \therefore AB>BC\text{ and }AB>AC.
\displaystyle {\therefore AB\text{ is the largest side of }\triangle ABC.}
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{In }\triangle ABC,\ D\text{ is any point on side }BC.
\displaystyle \text{Show that }AB+BC+AC>2AD.
\displaystyle \textbf{Answer:} \displaystyle \text{In }\triangle ABD,\text{ the sum of any two sides is greater than the third side.}
\displaystyle AB+BD>AD.\qquad\ldots\text{(I)}
\displaystyle \text{In }\triangle ACD,\text{ the sum of any two sides is greater than the third side.}
\displaystyle AC+CD>AD.\qquad\ldots\text{(II)}
\displaystyle \text{Adding (I) and (II), we get}
\displaystyle AB+BD+AC+CD>2AD.
\displaystyle \text{Since }D\text{ lies on }BC,\quad BD+CD=BC.
\displaystyle {\therefore AB+BC+AC>2AD.}
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{In }\triangle ABC,\ AC>AB.\text{ The internal bisector}
\displaystyle \text{of }\angle A\text{ meets the opposite side at }D.\text{ Prove that}
\displaystyle \angle ADC>\angle ADB.
\displaystyle \textbf{Answer:} \displaystyle \text{Since }AC>AB,\text{ the angle opposite }AC\text{ is greater than the angle}
\displaystyle \text{opposite }AB.
\displaystyle \therefore \angle ABC>\angle ACB.\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,\quad \angle ADC\text{ is an exterior angle.}
\displaystyle \therefore \angle ADC=\angle BAD+\angle ABD.
\displaystyle \text{In }\triangle ACD,\quad \angle ADB\text{ is an exterior angle.}
\displaystyle \therefore \angle ADB=\angle DAC+\angle ACD.
\displaystyle \text{Since }D\text{ lies on }BC,
\displaystyle \angle ABD=\angle ABC\text{ and }\angle ACD=\angle ACB.
\displaystyle \text{Using (I) and (II),}
\displaystyle \angle BAD+\angle ABD>\angle DAC+\angle ACD.
\displaystyle {\therefore \angle ADC>\angle ADB.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{In isosceles }\triangle ABC,\text{ sides }AB\text{ and }AC
\displaystyle \text{are equal. If }D\text{ lies on }BC\text{ and }E\text{ lies on }BC\text{ produced}
\displaystyle \text{through }C,\text{ prove that:}
\displaystyle \text{(i) }AC>AD\qquad\text{(ii) }AE>AC\qquad\text{(iii) }AE>AD.
\displaystyle \textbf{Answer:} \displaystyle \text{Since }AB=AC,\text{ the angles opposite these equal sides are equal.}
\displaystyle \therefore \angle ABC=\angle ACB.
\displaystyle \text{(i) Since }B,\ D\text{ and }C\text{ are collinear, }\angle ADC\text{ is an exterior angle}
\displaystyle \text{of }\triangle ABD.
\displaystyle \therefore \angle ADC>\angle ABD.
\displaystyle \text{Since }D\text{ lies on }BC,\quad \angle ABD=\angle ABC.
\displaystyle \text{Also, }\angle ACD=\angle ACB.
\displaystyle \text{But }\angle ABC=\angle ACB.
\displaystyle \therefore \angle ADC>\angle ACD.
\displaystyle \text{In }\triangle ACD,\text{ the greater angle has the greater side opposite to it.}
\displaystyle {\therefore AC>AD.}\qquad\ldots\text{(I)}
\displaystyle \text{(ii) Since }\angle ABC=\angle ACB\text{ and }\angle BAC>0^\circ,
\displaystyle 2\angle ACB<180^\circ.
\displaystyle \therefore \angle ACB<90^\circ.
\displaystyle \text{Since }BC\text{ is produced to }E,
\displaystyle \angle ACE=180^\circ-\angle ACB>90^\circ.
\displaystyle \text{In }\triangle ACE,\text{ the other two angles are each less than }90^\circ.
\displaystyle \therefore \angle ACE>\angle AEC.
\displaystyle \text{The greater angle has the greater side opposite to it.}
\displaystyle {\therefore AE>AC.}\qquad\ldots\text{(II)}
\displaystyle \text{(iii) From (I) and (II),}
\displaystyle AE>AC\text{ and }AC>AD.
\displaystyle {\therefore AE>AD.}
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Given }ED=EC.\text{ Prove that }AB+AD>BC. \displaystyle \textbf{Answer:}
\displaystyle \text{Join }B\text{ and }D.
\displaystyle \text{In }\triangle ABD,\text{ the sum of any two sides is greater than the third side.}
\displaystyle AB+AD>BD.
\displaystyle \text{Since }E\text{ lies on }BD,\quad BD=BE+ED.
\displaystyle \therefore AB+AD>BE+ED.
\displaystyle \text{But }ED=EC.
\displaystyle \therefore AB+AD>BE+EC.\qquad\ldots\text{(I)}
\displaystyle \text{In }\triangle BEC,
\displaystyle BE+EC>BC.\qquad\ldots\text{(II)}
\displaystyle \text{From (I) and (II),}
\displaystyle {\therefore AB+AD>BC.}
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{In }\triangle ABC,\ AB>AC\text{ and }D\text{ is a point}
\displaystyle \text{on side }BC.\text{ Show that }AB>AD.
\displaystyle \textbf{Answer:} \displaystyle \text{Since }AB>AC,\text{ the angle opposite }AB\text{ is greater than the angle}
\displaystyle \text{opposite }AC.
\displaystyle \therefore \angle ACB>\angle ABC.\qquad\ldots\text{(I)}
\displaystyle \text{Since }D\text{ lies on }BC,
\displaystyle \angle ACD=\angle ACB\text{ and }\angle ABD=\angle ABC.
\displaystyle \text{In }\triangle ACD,\quad \angle ADB\text{ is an exterior angle.}
\displaystyle \therefore \angle ADB>\angle ACD.
\displaystyle \therefore \angle ADB>\angle ACB.
\displaystyle \text{Using (I),}\quad \angle ADB>\angle ACB>\angle ABC=\angle ABD.
\displaystyle \therefore \angle ADB>\angle ABD.
\displaystyle \text{In }\triangle ABD,\text{ the greater angle has the greater side opposite to it.}
\displaystyle {\therefore AB>AD.}
\displaystyle \\

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