#### Question

In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.

#### Solution

`∠`ADB = `∠`ACB …… (i) (Angles in same segement)

Similarly ,

`∠`ABD = `∠`ACD …….. (ii)

But, `∠`ACB = `∠`ACD (AC is bisector of `∠`BCD)

∴ `∠`ADB = `∠`ABD (from (i) and (ii) )

TAS is a tangent and AB is a chord

∴ `∠`BAS = `∠`ADB (angles in alternate segment)

But, `∠`ADB = `∠`ABD

∴ `∠`BAS = `∠`ABD

But these are alternate angles

Therefore, TS ∥ BD

Is there an error in this question or solution?

Solution In a Cyclic Quadrilateral Abcd, the Diagonal Ac Bisects the Angle Bcd. Prove that the Diagonal Bd is Parallel to the Tangent to the Circle at Point A. Concept: Cyclic Properties.