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In the figure \( P O \perp Q 0 \). The tangents to the circle at \( P \) and \( Q \) intersect at a point \( T \). Prove that \( P Q \) and \( O T \) are right bisectors of each other.
Given:
In the figure \( P O \perp Q 0 \). The tangents to the circle at \( P \) and \( Q \) intersect at a point \( T \).
To do:
We have to prove that \( P Q \) and \( O T \) are right bisectors of each other.
Solution:
$PT$ and $QT$ are tangents to the circle.
$PT = QT$
$PO\ \perp\ QO$
$OP$ and $OQ$ are radii of the circle and $\angle POQ = 90^o$
$OQTP$ is a square where $PQ$ and $OT$ are diagonals.
Diagonals of a square bisect each other at right angles.
$PQ$ and $OT$ bisect each other at right angles.
Therefore, $PQ$ and $QT$ are right bisectors of each other.
Hence proved.