Question

Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the center

Answer 1

Given: $XY$ and $X^{\prime }{Y}^{\prime }$ are two parallel tangents to the circle wth centre $O$ and $AB$ is the tangent at the point $C$,which intersects $XY$ at $A$ and $X^{\prime }{Y}^{\prime }$ at $B$.

To Prove:$\angle AOB={90}^{\circ }$

Construction:Join $OC$

Proof:In $△OPA$ and $△OCA$

$OP=OC$ (radii)

$AP=AC$ (Tangents from point $A$)

$AO=AO$ (Common )

$△OPA\cong △OCA$ (By SSS criterion)

$\therefore ,\angle POA=\angle COA$ .... $\left(1\right)$ (By C.P.C.T)

$OQ=OC$ (radii)

$BQ=BC$ (Tangents from point $A$)

$BO=BO$ (Common )

$△OQB\cong △OCB$

$\angle QOB=\angle COB$ .......$\left(2\right)$

$POQ$ is a diameter of the circle.

Hence, it is a straight line.

$\therefore ,\angle POA+\angle COA+\angle COB+\angle QOB={180}^{\circ }$

From equations $\left(1\right)$ and $\left(2\right)$, it can be observed that

$2\angle COA+2\angle COB={180}^{\circ }$

$\therefore \angle COA+\angle COB={90}^{\circ }$

$\therefore \angle AOB={90}^{\circ }$