\displaystyle \textbf{Question 1: }\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 2: }\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 3: }\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 4: }\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 5: }\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 6: }\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 7: }\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 8: }\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 9: }\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 10: }\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 11: }\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 12: }\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 13: }\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 14: }\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 15: }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 16: }\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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