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 $\angle BCG={108}^o$ and $O$ is the centre of the circle, find:
(i) $\angle BCT$
(ii)$\angle DOC$

Answer 1

Join $OC,OD$ and $AC$

$\left(i\right)$

$\angle BCG+\angle BCD={180}^o$ [Linear pair]

$\Rightarrow {108}^o+\angle BCD={180}^o$

$\Rightarrow \angle BCD={180}^o-{108}^o={72}^o$

$BC=CD$

$\therefore \angle DCP=\angle BCT$

But,$\angle BCT+\angle BCD+\angle DCP={180}^o$

$\therefore \angle BCT+\angle BCT+{72}^o={180}^o$

$2\angle BCT={180}^o-{72}^o$

$\angle BCT={54}^o$

$\left(ii\right)$

PCT is tangent and CA is a chord

$\angle CAD=\angle BCT={54}^o$

But arc DC subtends $\angle DOC$ at the centre and $\angle CAD$ at the remaining part of the circle.

$\therefore \angle DOC=2\angle CAD=2\times {54}^o={108}^o$