Question:
In the given figure, AD ⊥ BC and CD > BD. Show that AC > AB.
Solution:
Given: AD ⊥ BC and CD > BD
To prove: AC > AB
Proof:
$\angle A D B=\angle A D C=90^{\circ} \quad(A D \perp B C) \quad \ldots(1)$
$\angle B A D<\angle D A C \quad(C D>B D) \quad \ldots(2)$
In $\Delta A B D$
Using angle sum property of a triangle,
$\angle B=180^{\circ}-\angle A D B-\angle B A D$
$\angle B=90^{\circ}-\angle B A D \quad \ldots(3)$
In $\Delta A D C$
Using angle sum property of a triangle,
$\angle A C D=90^{\circ}-\angle D A C \quad \ldots(4)$
From (2), (3) and (4), we get
$\angle B>\angle C$
Therefore, $A C>A B$.
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