In a triangle AD is bisector of angle BAC. Prove that AB>AD
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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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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.
hope it helped u a little.......☺️
follow to get instant answers.
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