Question

Prove that the tangents at the extremities of any chord make equal angles with the chord. [3 MARKS]

Answer 1

Concept: 1 Mark
Application: 2 Marks

Let AB be a chord of a circle with center O, and let AP and BP be the tangents at A and B respectively.

Suppose the tangents meet at P .

Join OP. Suppose OP meets AB at C . We have to prove that $\angle PAC=\angle PBC$ .

In triangles PCA and PCB , We have

PA = PB

$\angle APC=\angle BPC$
[The tangents are equally inclined to line joining extrernal point and centre of circle.]

PC = PC [Common]

So, by SAS - criterion of congruence, we have

$\Delta PAC\cong \Delta PBC$

$\Rightarrow \angle PAC=\angle PBC$

Alternative Solution:

In $\Delta PAB$, $PA=PB$ ( $\because$ tangents drawn from an external point to a circle are equal.)

$\because$ PA = PB, $\angle$PAB = $\angle$ PBA (Equal opposite sides make equal angles)

$\Rightarrow$ $\angle PAC$ = $\angle PBC$