Question

two circles intersect each other at point C and D. Their common tangent AB touches circle at point A and B. Prove that angle ADB + angle ACB =180°​

Answer 1

Angle ADB + Angle ACB = 180°​

Draw Segment CD

As per Tangent - Secant Angle Theorem

∠DAB = ∠ACD ------- (Eq.1)

∠DBA = ∠DCB ------- (Eq.2)

From Eq. 1 & Eq. 2

∠DAB + ∠DBA = ∠ACD + ∠DCB

Also, ∠ACB = ∠ACD + ∠DCB -----(Eq.3)

In Δ ADB,

⇒ ∠DAB + ∠DBA + ∠ADB = 180°

∴ ∠ACD + ∠DCB + ∠ADB = 180° (From Eq. 1&2)

∴ ∠ACB + ∠ADB = 180° (From Eq. 3)

Hence Proved