Question

ABC is a triangle in which $\angle B=2\angle C.$ D is a point on side BC such that AD bisects $\angle BAC$ and AB =CD. Find $m$ if $\angle BAC=m^{\circ }$

Answer 1

In $△BPC$, we have

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

Now, in $△ABP$ and $△DCP$, we have

$\angle ABP=\angle DCP=y$

$AB=DC$ --------------[Given]

and, $BP=PC$ ------[Using (1)]

So, by $SAS$ congruence criterion, we have

$△ABP\cong △DCP$

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

So in $△APD$, $AP=DP$

$\angle ADP=\angle DAP=x$

In $△ABD$, we have

$\angle ADC=\angle ABD+\angle BAD⟹3x=2y+x$

$x=y$

In $△ABC$, we have

$\angle BAC+\angle CBA+\angle ACB={180}^{\circ }$

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

$5x={180}^{\circ }$

$x={36}^{\circ }$

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

$\therefore \angle BAC=m^{\circ }={72}^{\circ }$