Question

Question 6
In figure AB and CD are common tangents to two circles of equal radii. Prove that AB = CD.

Answer 1

Givne AB and CD are tangents to two circle of equal radii
To prove AB =CD


Construction Join OA, OC, O’ B and O’D
Proof Now, $\angle OAB={90}^{\circ }$
[ Tangents at any point of a circle is perpendicular to radius through the point of contact]
Thus, AC is a straight line
Also. $\angle OAB+\angle OCD={180}^{\circ }$
$\therefore AB\parallel CD$
Similarly, BD is straight line
And $\angle O^{\prime }BA=\angle O^{\prime }DC={90}^{\circ }$
Also, AC = BD [ radii of two circles are equal]
In quadrilateral ABCD. $\angle A=\angle B=\angle C=\angle D={90}^{\circ }$
And AC =BD
Hence. AB=CD [ opposite sides of rectangle are equal]