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 =1${08}^o$ and O is the centre of the circle, find :

(i) angle BCT

(ii) angle DOC

Answer 1

BC=CD (given)

$\angle BCG=108°$
Construction: B and D are joined.OC and OD are joined.

To find: ∠DOC and ∠BCT

as $\angle BCG+\angle BCD=180°$ (linear pair)

So
$angleBCD=180°-108°=72°$

$\angle CBD=\angle CDB=x°$(as BC=CD, △BCD is isosceles, so two angles are equal)
In △BCD,
$\angle CBD+\angle CDB+\angle BCD=180°$

x+x+72°=180°
2x=180−72

x=54°
and $\angle BCT=\angle CDB$ (angle −tangent theorem)

as
$\angle CDB=54°$
so
$\angle BCT=54°\&\angle CBD=54°$

$\angle DOC=2\times \angle CBD$( as angle at the center is 2 times of angle at cicumference on the same chord)

so
$\angle DOC=2\times 54°=108°$
Hence $\angle DOC=108°$ and

$\angle BCT=54°$