#### 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)

Is there an error in this question or solution?

Solution 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° Concept: Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers.