ICSE Class X Math 2015 Paper

$A B C$ is a right angled triangle with $\angle A B C=$ $90^{\circ} \cdot D$ is any point on $A B$ and $D E$ is perpendicular to AC. Prove that :

Question : 47 of 52

Marks: +1, -0

\triangle A D E \sim \triangle A C B .

Solution:

To Prove :

\triangle ADE \sim \triangle ACB

(i) In

\triangle ADE

and

\triangle ACB

\;\text{(1)}\;\; \angle A\;=\angle A\;\; \text{[common]}\;

(2)

\angle AED\;=\angle ABC

\;=90^{\circ}\;\text{(given)}\;

\therefore \triangle ADE \sim \triangle ACB

[AA axiom]

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