Given: A ∆ABC in which AD is the bisector of ∠A meeting BC in D such that BD = CD
To Prove: ∆ACD is an isosceles triangle.
Construction: We produce AD to E such that AD = DE and join EC.
Proof: In ∆ADB and ∆EDC we have
AD = DE [By construction]
∠ADB = ∠CDE [Vertically opposite angles]
BD = DC [Given]
∴ By SAS criterion of congruence, we get
∆ADB ≅ ∆EDC
⇒ AB = EC ...(i) And, ∠BAD = ∠CED [By cpctc]
But, ∠BAD = ∠CAD
∴ ∠CAD = ∠CED
⇒ AC = EC [Sides opposite to equal angles are equal]
⇒ AC = AB [By eg. (i)]