Question

In the adjoining figure, $O$ is the centre of the circle and $AB$ is the diameter. At point $C$ on the circle, the tangent $CD$ is drawn. Line $BD$ is the tangent to the circle at point $B$. Show that seg $OD\parallel$ chord $AC$.

Answer 1

Given : $O$ is the centre of the circle.
Seg $AB$ is a diameter, $CD$ is the tangent at $C$.
$BD$ is a tangent at $B$
In $\Delta COD$ and $\Delta BOD$
$\Rightarrow OC=OB$ ....(Radii of same circle)
$\Rightarrow CD=DB$ ....(Tangent from an external point)
$\Rightarrow OD=OD$ ....(Common side)
$\Rightarrow \Delta COD=\Delta BOD$ (SSS test)
$\Rightarrow \Delta COD=\Delta BOD=x$ (c.a.s.t.) equation (i)
In $\Delta AOC$, $AO=OC$ ....(Radii of same circle)
$\Rightarrow$ $\angle OAC=\angle OCA=y$ ....(Isoceles $\Delta$ theorem) equation (ii)
$\Rightarrow \angle COB=\angle OAC+\angle OCA$ ....(Remote interior $\angle$ theorem)
$\Rightarrow \angle COD+\angle BOD=\angle OAC+\angle OCA$
$\Rightarrow x+x=y+y$ ....[From equation (i) and (ii)]
$\Rightarrow 2x=2y$
$\Rightarrow x=y$ ....(dividing by $2$)
$\Rightarrow \angle COD=\angle OCA$
$\Rightarrow$ Seg $OD\parallel$ chord $AC$. ....(Alternate $\angle s$ test)

Hence, proved.