Question

ABCD is a cyclic quadrilateral and the tangent at A to the circle circumscribing the quadrilateral is parallel to BD. Prove that CA is a bisector of angleDCB​

Answer 1

Answer:

DAB = 90° (Angle in a semi-circle)

In ΔABD,

∠ABD + ∠ADB + ∠DAB = 180°

⇒ ∠ADB = 180° – (60° + 90°)

⇒∠ADB = 180° – 150°

⇒∠ADB = 30°

∠DCQ = ∠CBD = 40°

(Alternate segment theorem)

∠BCD = 90°

In ΔBCD,

∠BCD + ∠BDC + ∠DBC = 180°

⇒ ∠BDC = 180° – (90° + 40°) = 50°

∠BDC = ∠BCP = 50° (Alternate segment theorem)

∴ (i) ∠ADB = 30°

(ii) ∠BCP = 50°

Step-by-step explanation:

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