1.

Then BDDC

Answer»

AD^2 = BD × CD

AD ÷ CD = BD ÷ AD

ΔADC ∼ ΔBDA (by SAS ∠D = 90°)

∠BAD = ∠ACD ;

∠DAC = ∠DBA (Corresponding angles of similar triangles)

∠BAD+ ∠ACD + ∠DAC + ∠DBA = 180°

⇒ 2∠BAD + 2∠DAC = 180°

⇒ ∠BAD + ∠DAC = 90°

∴ ∠A = 90°


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