In the given figure AE is bisector of - BAC and of - BDC- show that - ABD - ACD and hence BD - CD

∠ BAD = ∠ CAD, AD = AD, ∠ BDA = ∠ CDA ⇒ By ASAcongruency; ABD ≅ ACD, by c.p.c.t. BD = CD . ...

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Properties of Isosceles and Equilateral Triangles

In the given ...

Question

In the given figure AE is bisector of $∠ B A C$ and of $∠ B D C$ . show that $Δ A B D ≅$ ACD and hence BD = CD

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∠ B A C

A D ⊥ B C

∠ D A E = \frac{1}{2} (∠ C - ∠ B)

∠ A B D = ∠ B D C

C D = 4 A B

B D = 5 B E

In the given figure, D is a point on side BC of ΔABC such that. Prove that AD is the bisector of ∠BAC.

Δ A B C

Δ A B D ≅ Δ A E

Δ A B D ≅ Δ A C E

Δ A B C

Δ A B D ≅ Δ A E

Δ A B D ≅ Δ A C E

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