Question

In ΔABC, AE is the bisector of ∠BAC. D is a point on AC such that AB = AD. Prove that BE = DE and ∠ABD > ∠C.

Answer 1

In the ΔABE and ΔADE,
AB = AD   ....(Given)
∠BAE = ∠DAE   ....(AE is the bisector of ∠BAC)
AE = AE     ....(Common side)
∴ ΔABE ≅ ΔADE   ....(SAS test)
⇒ BE = DE   ....(c.p.c.t.c)
In ΔABD,
AB = AD
⇒ ∠ABD = ∠ADB
∠ADB > ∠C   ...(Exterior angle property)
⇒ ∠ABD > ∠C.