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Question
ABC is a right angled triangle with ∠ABC = 90°. D is any point on AB and DE is perpendicular to AC. Prove that :
ΔADE ~ ΔACB.
Solution
In ΔADE and ΔACB.
∠AED = ∠ABC ...(both are right angles)
∠DAE = ∠CAB ....(common angles)
ΔADE ∼ ΔACB ...(AA criterion for similarity)
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