Question

In the above figure, $AB=AC$ and $D$ is the mid-point of $BC$
Prove that (i) Is $△ADB\cong △ADC$? (Give reason)
(ii) Is $\angle B=\angle C$? Why?

Answer 1

Given: $AB=AC$ and D is midpoint

So, $BD=DC\dots \dots \left(i\right)$

In $\Delta ABD\&\Delta ACD$, we have

$AB=AC\left(given\right)$

$BD=DC\left(fromeq\right(i\left)\right)$

$AD=AD\left(common\right)$

$\left(i\right)\Delta ADB\cong \Delta ADC$ [$BySSScongruencyrule]$

$\left(ii\right)\angle B=\angle C\left(CPCT\right)$