Question

ABC is a triangle in which $\angle B=2\angle C$. D is a point on BC such that AD bisects $\angle BAC$ and $AB=CD.$ Find $\angle BAC$

• A. ${72}^0$
• B. ${36}^0$
• C. ${108}^0$
• D. ${90}^0$

Answer 1

The correct option is A ${72}^0$

In $△ABC$, we have,

$\angle B=2\angle C$ or $\angle B=2y$ where $\angle C=y$.

$AD$ is the bisector of $\angle BAC$. So, let $\angle BAD=\angle CAD=x$.

Let $BP$ be the bisector of $\angle ABC$.

In $△BPC$, we have,

$\angle CBP=\angle BCP=y⟹BP=PC$ ........(1)

Now, in $△ABP$ and $△DCP$, we have,

$\angle ABP=\angle DCP=y$

$AB=DC$

and, $BP=PC$

So, by SAS congruence criterion,

$△ABP\cong △DCP$.

Therefore, $\angle BAP=\angle CDP=2x$ and $AP=DP$

$⟹\angle ADP=\angle DAP=x$

In $△ADB$, we have,

$\angle ADC=\angle ABD+\angle BAD$

$⟹3x=2y+x$

$x=y$

In $△ABC$, we have,

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

$2x+2y+y={180}^o$

$5x={180}^o$

$x={36}^o$

Hence, $\angle BAC=2x={72}^o$