Question

In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that:
(i) BD = CD
(ii) ED = EF

Answer 1

(i) AB = AC
This means that $\Delta ABC$ is isosceles. In an isosceles triangle, median & altitude are same. Thus, AD is median for side BC.
$\Rightarrow$ BD = DC (or CD)
(ii) In $\Delta$ EFB & EDB
$\angle EFB=\angle EDB(={90}^{\circ })$ ---- (1)
$\angle EFB=\angle EBD$ (BE is the angle bisector of $\angle B$) ---- (2)
BE = BE (common)
Thus, by AAS congruency condition
$\Delta EFB\cong \Delta EDB\Rightarrow EF=ED\left(CPCT\right)$