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Question
In △ABD, C is a point on side BD such that ∠ACD = ∠BAD. Is △BAD similar to triangle ACD? If yes, then which axiom is satisfied:
Options
Yes, ASA
Yes, SAS
Yes, AA
No
MCQ
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Solution
Yes, AA
Explanation:
From figure,
In △ BAD and △ ACD,
⇒ ∠BAD = ∠ACD (Given)
⇒ ∠ADB = ∠ADC (Common angle)
∴ △ BAD ~ △ ACD (By A.A. axiom)
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