Question

in figure AB=AD,AcAE and <BaD=<EAc that prove that Bc=DE​

Answer 1

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)