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:

विकल्प

• Yes, ASA

• Yes, SAS

• Yes, AA

• No

Answer 1

Yes, AA

Explanation:

From figure,

In △ BAD and △ ACD,

⇒ ∠BAD = ∠ACD (Given)

⇒ ∠ADB = ∠ADC (Common angle)

∴ △ BAD ~ △ ACD (By A.A. axiom)