Question

Prove that the lengths of tangents drawn from an external point to a circle are equal.

Answer 1

Hint: First draw a circle with two tangents touching it with the same external point then join the points touching the circle and the center and atlast join the center of the circle to the external point, you will get two triangles trying to prove them congruent to get the required result.

Complete step by step answer:

First of all draw the figure as stated in the hint and then we will examine the facts

Given: A circle with centre O; PA and PB are two tangents to the circle drawn from an external point P.

To prove: PA = PB

Construction: Join OA, OB, and OP.

It is known that a tangent at any point of a circle is perpendicular to the radius through the point of contact.

$OA \bot PA$

$OB \bot PB$

In triangle OPA and OPB

$\angle OPA = \angle OPB$

$OA = OB(radii)$

$OP = OP(common)$

Therefore triangle OPA is congruent to triangle OPB by RHS criterion.

Which means \[PA = PB\]

(Corresponding parts of congruent triangles are equal)

Thus, it is proved that the lengths of the two tangents drawn from an external point to a circle are equal.

The length of tangents drawn from any external point are equal.

Note:

RHS criterion for Congruence states that when a triangle has one right angle and the hypotenuse is equal and also any one of the sides other than hypotenuse is also equal then they are considered as congruent triangles by RHS criterion. Also if two triangles are congruent then all the corresponding sides and angles are equal.