#### Question

In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠BCG = 108° and O is the centre of the circle, find:

(i) Angle BCT

(ii) angle DOC

#### Solution

Join OC, OD and AC

i)

`∠`BCG + `∠`BCD = 180° (Linear pair)

⇒ 180 °+ `∠`BCD = 180°

⇒ `∠`BCD = 180° -180° =72°

BC = CD

∴ `∠`DCP = `∠`BCT

But, `∠`BCT + `∠`BCD + `∠`DCP = 180°

∴ `∠`BCT + `∠`BCT + 72° = 180°

2`∠`BCT = 180° - 72°

`∠`BCT = 54°

ii)

PCT is a tangent and CA is a chord.

`∠`CAD = `∠`BCT = 54°

But arc DC subtends `∠`DOC at the centre and `∠`CAD at the

remaining part of the circle.

∴ `∠`DOC = 2`∠`CAD = 2 × 54° = 108°

Is there an error in this question or solution?

Solution Abcd is a Cyclic Quadrilateral with Bc = Cd. Tc is Tangent to the Circle at Point C and Dc is Produced to Point G. If ∠Bcg = 108° and O is the Centre of the Circle Find: (I) Angle Bct (Ii) Angle Doc Concept: Cyclic Properties.