Question

ABC is a triangle in which $\angle B=2\angle C.D$ is a point on BC such that AD bisects $\angle BACandAB=CD.\text{Prove that}\angle BAC={72}^{\circ }.$

Answer 1

Given : In $\Delta ABC,$

$\angle B=2\angle C,AD\text{is the bisector of}\angle BAC$

AB = CD

To prove : $\angle BAC={72}^{\circ }$

Donstruction : Draw bisector of $\angle B$ which meets AD at O

Proof : AC at E join ED

In $\Delta ABC,$

$\angle B=2\angle C$

OB is the bisector of $\angle B$

$\therefore \angle C=\frac{1}{2}\angle B=\angle 1$

$\therefore In\Delta BCE,$

$\angle 1=C$

$\therefore BE=EC\text{(Opposite side of equal angles)}$

Now in $\Delta ABEand\Delta CDE,$

AB = CD (Given)

$\angle 2=\angle C$

$BE=EC$

$\therefore \Delta ABE\cong \Delta CDE\left(SASaxiom\right)$

$\therefore \angle BAE=\angle CDE(c.p.c.t.)$

$\therefore AE=ED\left(Oppositesides\right)$

Let $\angle CDE=2x,and\angle ADE=\angle DAE=x$

$(\because \angle A=2x)$

In $\Delta$ ABD, we have,

$\angle ADC=\angle ABD+\angle BAD\Rightarrow x+2x=2y+x$

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

In $\Delta ABC,$

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

$\left(Anglesofatriangle\right)$

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

$\Rightarrow 2x+x+x={180}^{\circ }\Rightarrow 5x={180}^{\circ }$

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

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

Hence proved.