Question

In the given figure, $AB$ is a chord of a circle with center $O$ and $AB$ is produced to $C$ such that $BC=OB$. Also, $CO$ is joined and produced to meet the circle in $D$. If $\angle ACD=y^o$ and $\angle AOD=x^o$, prove that $x=3y$

Answer 1

It is given that $AB$ is a chord of a circle with centre $O$ ans $AB$ is produced to $C$ such that $BC=OB$

We know that $\angle BOC$ and $\angle BCO$ form isosceles triangle

$\angle BOC=\angle BCO=y^o$

We know that $OA$ and $OB$ are the radii of same circle

$OA=OB$

We know that $\angle OAB$ and $\angle OBA$ form isosceles triangle

$\angle OAB=\angle OBA=2y^o$

From the figure we know that

Exterior $\angle AOD=\angle OAC+\angle ACO$

It can be written as

Exterior $\angle AOD=\angle OAB+\angle BCO$

So we get

Exterior $\angle AOD=3y^o$

It is given that

$\angle AOD=x^o$

So we get

$x^o=3y^o$

Therefore it is proved that $x=3y$