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Prove that If the bisector of the vertical angle bisects the base of the triangle, then the triangle is isosceles.  

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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)] 

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