Q6 of 45 Page 49

In figure 3.61, bisector of ∠BAC intersects side BC at point D.

Prove that AB > BD


Given: AD is bisector of ∠BAC


To Prove: AB > BD


Proof: AD is bisector of ∠BAC


⇒ ∠BAD = ∠DAC …..(1)


Now, In ΔADC, ∠ ADB is the exterior angle


∠ADB > ∠DAC ..(2) [exterior angle of a triangle is greater than each


of the opposite interior angles]


Substituting ∠DAC = ∠BAD in (2)


⇒ ∠ADB > ∠BAD


⇒ AB > BD [side opposite to larger angle is larger]


More from this chapter

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4

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In figure 3.63, seg AD ⊥ seg BC. seg AE is the bisector of ∠CAB and C - E - D. Prove that ∠DAE = 1/2 (∠B - ∠C)