In the adjoining figure, ∆ABC is an isosceles triangle in which AB = AC. Also, D is a point such that BD = CD.
Prove that AD bisects ∠A and ∠D

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. $Given : △ A B C is an isosceles triangle . A B = A C B D = C D To prove : A D bisects ∠ A and ∠ D . Proof : Consider △ A B D and △ A C D : A B = A C (given) BD = CD (given) AD = AD (common) B y S S S congruence property : △ A B D ≅ △ A C D = > ∠ B A D = ∠ C A D (by cpct) = > ∠ B D A = ∠ C D A (by cpct)$

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In the adjoining figure, ∆ABC is an isosceles triangle in which AB = AC. Also, D is a point such that BD = CD. Prove that AD bisects ∠A and ∠D Figure

.Given:△ABC is an isosceles triangle.AB=ACBD =CD To prove:AD bisects ∠A and ∠D.Proof:Consider △ABD and △ACD:AB=AC (given)BD=CD (given)AD=AD (common) By SSS congruence property:△ABD≅ △ACD =>∠BAD=∠CAD (by cpct)=>∠BDA=∠CDA (by cpct)

In the adjoining figure, ∆ABC is an isosceles triangle in which AB = AC. Also, D is a point such that BD = CD.
Prove that AD bisects ∠A and ∠D

Figure