1.

In figure, AD is bisector of ∠BAC then prove that AB > BD.

Answer»

In ∆ABC

AD is the bisector of ∠BAC

⇒ ∠1 = ∠2 …(i)

Also in ∆ADC

∠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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