Question

Question 6
ABC is a right triangle with AB = AC. If bisector of $\angle A$ meets BC at D, then prove that BC = 2AD.

Answer 1

$\Delta ABC$ is a right angled triangle with AB = AC, AD is the bisector of $\angle A$.


$In\Delta ABC,AB=AC\left[given\right]\Rightarrow \angle C=\angle B...\left(i\right)$
[angles opposite to equal sides are equal]
Now, in right angled $\Delta ABC,\angle A+\angle B+\angle C={180}^{\circ }$
[angle sum property of a triangle]
$\Rightarrow {90}^{\circ }+\angle B+\angle B={180}^{\circ }[fromEq.(i\left)\right]\Rightarrow 2\angle B={90}^{\circ }\Rightarrow 2\angle B={90}^{\circ }\Rightarrow \angle B={45}^{\circ }\Rightarrow \angle B=\angle C={45}^{\circ }or\angle 3=\angle 4={45}^{\circ }Now,\angle 1=\angle 2={45}^{\circ }[\because ADisbisectorof\angle A]\therefore \angle 1=\angle 3,=\angle 2=\angle 4\Rightarrow BD=AD,DC=AD..........\left(ii\right)$
[sides opposite to equal angles are equal]
Hence, BC = BD + CD = AD + AD
$BC=2AD[fromEq.(ii\left)\right]$