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Question

In the given figure, AB and CD are common tangents to two circles of unequal radii. if radii of the two circles are equal ,prove that AB=CD.
427341_9ad06772139b4563bfb8c840a2f7b105.png


Solution

Given : Two circles of equal radii, two common tangents, AB and CD on circles, $$C_1$$ and $$C_2$$.

To prove : $$AB = CD$$

Construction : Join $$O_1 A, O_1 C$$ and $$O_2 B$$ and $$O_2D$$ . Also join $$O_1 O_2$$.

Proof : Since tangent at any point of a circle is perpendicular to the radius to the point of contact.

$$\therefore \angle O_1 AB = \angle O_2 BA  = 90^{\circ}$$

As $$O_1A = O_2B$$, so $$O_1 ABO_2$$ is a rectangle

Since opposite sides of a rectangle are equal 

$$\therefore AB = O_1 O_2 $$ ___(i)

Similarly, we can prove that $$O_1 CDO_2$$ is a rectangle.

$$\therefore O_1O_2 = CD$$ ___(ii)

From (i) and (ii) , we get

$$AB = CD$$

Hence proved.

1790893_427341_ans_ffad5d8397de4f46b4c8632fb3c2c49e.png

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