# Two direct common tangents are drawn to two non-intersecting circles.prove that the segments between the points of contact are equal.

We form our diagram from given information , As :

Here Point of contact of two direct chords is A , B and C , D .

We know " A tangent to a circle is perpendicular to the radius at the point of tangency. "

And we join center of " M " to tangents that meet at A and D , and another circle with center " N " to tangent that meet at " B " and " C " .

So,

$\angle $ MAB = 90$\xb0$ ,

$\angle $ MDC = 90$\xb0$

And

$\angle $ NBA = 90$\xb0$ ,

$\angle $ NCD = 90$\xb0$

SO,

In quadrilateral ABCD , all four angles are at 90$\xb0$ , So ABCD is a rectangle .

And We know opposite side of rectangle are equal to each other .

So,

**AB = CD ( Hence proved )**

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