Question

In the figure, ABC is a triangle in which $\angle B=2\angle C.\text{D is a point on side BC such that AD bisects}\angle BACandAB=CD.BE\text{is the bisector of}\angle B.\text{The measure of}\angle BACis$

• A. ${72}^{\circ }$
• B. ${73}^{\circ }$
• C. ${74}^{\circ }$
• D. ${95}^{\circ }$

Answer 1

The correct option is A

${72}^{\circ }$

In the figure, $\angle B=2\angle C,AD\text{and BE are the bisectors of}\angle Aand\angle B\text{respectively, AB=CD}$

$Let\angle C=x,then\angle B=2xand$

$Let\angle A=2y$

$\therefore \angle ABE=\angle CBE=x$

and $\angle BAD=\angle CAD=y$

$In\Delta BCE,$

$\therefore \angle EBC=\angle BCE=x$

$\therefore BE=EC$

$In\Delta ABEand\Delta DCE,$

$AB=DC\left(Given\right)$

$\angle ABE=\angle C(Each=x)$

$BE=EC\left(Proved\right)$

$\therefore \Delta ABE\cong \Delta DCE\left(\text{SAS axiom}\right)$

$\therefore \angle BAE=\angle EDC=2y$

$andAE=ED$

$\therefore \angle EAD=\angle EDA=y$

$In\Delta ABD,$

$Ext.\angle ADC=2x+y=2y+y$

$\Rightarrow 2x=2y\Rightarrow x=y$

$Nowin\Delta ABC,$

$\angle A+\angle B+\angle C={180}^{\circ }$

$\Rightarrow 2y+2x+x={180}^{\circ }\Rightarrow 2x+2x+x={180}^{\circ }$

$\Rightarrow 5x={180}^{\circ }\Rightarrow x=\frac{{180}^{\circ }}{5}={36}^{\circ }$

$\therefore \angle BAC=2y=2x=2\times {36}^{\circ }={72}^{\circ }$