Question

Prove that a diameter $AB$ of a circle bisects all those chords which are parallel to the tangent at the point $B.$

Answer 1

Consider a circle with center at $O.$ ${}^{\prime }{l}^{\prime }$ is tangent that touches the circle at point B. Let $AB$ is the diameter. Consider a chord $CD$ parallel to $l.$ $AB$ intersects $CD$ at point $Q.$

To prove the given statement, it is required to prove CQ = QD

Since $l$ is tangent to the circle at point B.

$\therefore AB\perp l$

It is given that $l\parallel CD$

Since $CD$ is a chord of the circle and $OB$ is radius.

So, Radius $OB$ is perpendicular tangent $l,$ $OB\perp l$ [radius from the center of the circle to the point of tangency is perpendicular to the tangent]

$\therefore CD\parallel l,$ so $OQ\perp CD.$

Thus, $OQ$ bisects the chord $CD.$ [The perpendicular from the center of a circle to a chord bisects the chord.]

This means $AB$ bisects chord $CD.$ at point $Q$