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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.
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Solution
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.
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