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 :

1. angle BCT
2. angle DOC

Answer 1

Join OC, OD and AC

i. ∠BCG + ∠BCD = 180°  ...(Linear pair)

`=>` 108° + ∠BCD = 180°   ...(∵ ∠BCG = 108° given)

`=>` ∠BCD = 180° – 108° = 72°

BC = CD   ...(Given)

∴ ∠DCP = ∠BCT

But, ∠BCT + ∠BCD + ∠DCP = 180°

∴ ∠BCT + ∠BCT + 72° = 180°  ...(∵ ∠DCP = ∠BCT)

2∠BCT = 180° – 72° = 108°

∴ ∠BCT = `108^circ/2` = 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°