In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
It is given that ∠BAD = ∠EAC
∠BAD + ∠DAC = ∠EAC + ∠DAC
∠BAC = ∠DAE
In ΔBAC and ΔDAE,
AB = AD (Given)
∠BAC = ∠DAE (Proved above)
AC = AE (Given)
ΔBAC ≅ ΔDAE (By SAS congruence rule)
BC = DE (By CPCT)
