In a triangle ABC if AD is the bisector of angle A show that AB >BD, AC>DC.

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Solution

If AD is the bisector of ∠A of triangle ABC, show that AB>DB.

Given :AD is the bisector of ∠A of triangle ABC .

Hence, ∠DAB=∠DAC
∠BDA is the exterior angle of the ∆DAC.

Hence, ∠BDA> ∠DAC
or ∠BDA > ∠DAB
Since ,∠DAC=∠DAB

→ AB > BD In a triangle sides opposite to greater angle is greater.

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Similar questions

In triangle ABC if AD is bisector of angle A show that AB>BD and AC > DC

△ A B C

A D

∠ A

A B = 8 c m, B D = 5 c m

D C = 4 c m

A C

In $△$ ABC, AB = 3 and, AC = 4 cm and AD is the bisector of $∠$ A. Then, BD : DC is —

△ A B C, A B = 3 c m, A C = 4 c m

∠

B D : D C

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