In an isosceles triangle $ABC$ with $AB = AC$,$D$ and $E$ are points on $BC$ such that $BE = CD$ (see Fig). Show that $AD = AE$.
(N/A) In $\Delta ABD$ and $\Delta ACE$:
$AB = AC$ (Given) ........... $(1)$
$\angle B = \angle C$ (Angles opposite to equal sides) ........... $(2)$
Also,$BE = CD$
Subtracting $DE$ from both sides:
$BE - DE = CD - DE$
$BD = CE$ ........... $(3)$
Therefore,$\Delta ABD \cong \Delta ACE$ (Using $(1)$,$(2)$,$(3)$ and $SAS$ congruence rule).
This gives $AD = AE$ $(CPCT)$.
$AB = AC$ (Given) ........... $(1)$
$\angle B = \angle C$ (Angles opposite to equal sides) ........... $(2)$
Also,$BE = CD$
Subtracting $DE$ from both sides:
$BE - DE = CD - DE$
$BD = CE$ ........... $(3)$
Therefore,$\Delta ABD \cong \Delta ACE$ (Using $(1)$,$(2)$,$(3)$ and $SAS$ congruence rule).
This gives $AD = AE$ $(CPCT)$.
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