$P$ is a point on the circle. A line is drawn through the center parallel to the tangent at $P$ and cutting the circle at $Q$. Prove that the tangent at $Q$ is perpendicular to the tangent at $P$.
Let $C$ be the center of the circle and $P$ any point on the circle.
$PA$ is the tangent at point $P$.
$CB$ is a line parallel to $PA$ and cutting the circle at the point $Q$.
$QD$ is the tangent to the circle at point $Q$.
The tangent to the circle at point $Q$ intersects $PA$ at the point $M$.
To prove that $\;$ $QD \perp PA$
i.e. $\;$ To prove that $\;$ $QM \perp PM$
From the figure,
$MP$ and $MQ$ are tangents drawn from the external point $M$.
$\therefore \;$ $MP = MQ$ $\;\;\; \cdots \; (1)$
$CP = CQ$ $\;\;\; \cdots \; (2)$ are the radius of the circle.
$\therefore \;$ $\angle CPM = 90^\circ$ $\;\;\; \cdots \; (3)$ $\;$ [radius is perpendicular to the tangent]
and $\;$ $\angle CQM = 90^\circ$ $\;\;\; \cdots \; (4)$
$\therefore \;$ From equations $(1), \; (2), \; (3) \;$ and $(4)$
$PMQC$ $\;$ is a square.
$\implies$ $\angle PMQ = 90^\circ$
i.e. $\;$ The tangent at $Q$ is perpendicular to the tangent at $P$.
Hence proved.