Question

Question 9
Prove that the tangents drawn at the ends of a chord of a circle make equal angles with the chord.

Answer 1

To prove : $\angle 1=\angle 2$.
Let RQ be a chord of the circle.
Tangents are drawn at the point R and Q.

Let P be another point on the circle, then.
Join PQ and PR.
Since, at point Q , there is a tangent.
$\therefore$ $\angle 2$ = $\angle$P [Angles in alternate segments are equal]
Since at point R, there is a tangent.
$\therefore$ $\angle 1$ = $\angle$P [Angles in alternate segments are equal]
$\therefore$ $\angle 1$ = $\angle$ 2 = $\angle$P
Hence proved.

Alternative Solution:

Extend the two tangents to a point O , where they meet
Something like this:
Here , RQ is the chord from whose ends the tangents are drawn.

The Tangents QO and PO meet at O. We need to prove
$\angle OQR=\angle ORQ$.

We know, OQ = OR
(Tangents drawn from an external point to the circle are equal)

$\therefore$ ,
$\angle OQR=\angle ORQ$ (Angles opposite to opposite sides are equal)
Hence , proved