$In the given figure A D ⊥ B C a n d C D > B D . S h o w t h a t A C > A B .$

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Solution

Given: AD ⊥ BC and CD > BD
To prove: AC > AB
Proof:
∠ADB=∠ADC=90°
AD⊥BC -----1

∠BAD<∠DAC

CD>BD -----2

In ∆ABD,
Using angle sum property of a triangle,
∠B=180°-∠ADB-∠BAD∠B=90°-∠BAD-----3
In ∆ADC,
Using angle sum property of a triangle,
∠ACD=90°-∠DAC ----- 4
From (2), (3) and (4), we get
∠B>∠C
Therefore, AC>AB.

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