in figure IF AD is a bisector of angle A show that AB is greater than BD and AC is greater than CD
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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.
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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Step-by-step explanation:
In ΔABC
AD is the bisector of ∠BAC
∠1=∠2 …(i)
Also in ΔADC
∠3=∠2+∠C (ext. angle is equal to sum of opposite interior angles)
∠3>∠2 (ext. angle is greater than one of the interior angles)
But ∠1=∠2
∠3>∠1
AB>BD
Hence proved.
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