Question

In the figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Prove that BC = DE.

Answer 1

In ΔADE and ΔBAC
AE = AC
AB = AD
∠BAD = ∠EAC
∠DAC = ∠DAC = DAC   ...(common)
⇒ ∠BAC = ∠EAD = EAD
Therefore, ΔADE ≅ ΔBAC  ...(SAS criteria)
Hence, BC = DE.