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