Question

in the figure, bisector of angle BAC intersects side BC at point D. Porve that AB > BD

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whoever will answer it in steeps and properly will get 30 pts reward​

Answer 1

$answer$

Given: Bisector of ∠BAC intersects side BC at point D.

To prove: AB > BD

Proof: ∠BAD ≅ ∠DAC ….(i) [Seg AD bisects ∠BAC]

∠ADB is the exterior angle of ∆ADC.

∴ ∠ADB > ∠DAC ….(ii) [Property of exterior angle]

∴ ∠ADB > ∠BAD ….(iii) [From (i) and (ii)]

In ∆ABD, ∠ADB > ∠BAD [From (iii)]

∴ AB > BD [Side opposite to greater angle is greater]

Answer 2

Given ,

<BAC intersects side BC at point D.

<DAC = <BAD ---( 1 )

Proof :

In ∆ADC , CD is extended to B.

exterior angle ADB > angle DAC

=> <ADB > angle BAD [ from ( 1 ) ]

AB > BD

[ Since ,

The sum of any two sides of a triangle

is greater than the third side . ]