$ A B $ and $ C D $ are common tangents to two circles of equal radii. Prove that $ A B=C D $.

Given:

\( A B \) and \( C D \) are common tangents to two circles of equal radii.

To do:
We have to prove that \( A B=C D \).
 Solution:

Join $OA, OC, OB$ and $OD$.

From the figure,

$\angle OAB = 90^o$    [Tangent at any point of a circle is perpendicular to the radius through the point of contact)

This implies,

$AC$ is a straight line.

$\angle OAB + \angle OCD = 180^o$

$AB\ \parallel\ CD$

Similarly,

$BD$ is a straight line and $\angle OBA = ∠ODC = 90^o$

$AC = BD$     (Radii of two circles are equal)

In quadrilateral $ABCD$,

$\angle A = \angle B = \angle C = \angle D = 90^o$

$AC = BD$

$ABCD$ is a rectangle.

Therefore,

$AB = CD$    (Opposite sides of a rectangle are equal)

Hence proved.

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Updated on: 10-Oct-2022

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