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Question
Two circles intersect each other at points C and D. Their common tangent AB touches the circles at point A and B. Prove that :
∠ ADB + ∠ ACB = 180°
Solution
Draw seg CD.
∠ DAB = ∠ ACD .... (1) Tangent secant angle theorem
∠ DBA = ∠ DCB .... (2) Tangent secant angle theorem
From (1) and (2)
∠ DAB + ∠ DBA = ∠ ACD + ∠ DCB
Now, ∠ ACB = ∠ ACD + ∠ DCB ...... (3)
In Δ ADB,
∠DAB + ∠ DBA + ∠ ADB = 180°.... (Sum of angles of a triangle.)
∴ ∠ACD + ∠DCB + ∠ADB = 180° ......From (1) and (2)
∴ ∠ACB + ∠ADB = 180° ......................From (3)
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