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Question

In figure, AB and CD are common tangents to two circles of unequal radii. Prove that AB$$=$$CD.
1603132_3104f94696a34ef8b031bf856461940f.png


Solution

Given : Circles $$C_1$$ and $$C_2$$ of radii $$r_1$$ and $$r_2$$ respectively and $$r_1 < r_2$$.


$$AB$$ and $$CD$$ are two common tangents.


To prove : $$AB = CD$$


Construction : Produce $$AB$$ and $$CD$$ upto P where both tangents meet.


Proof : Tangents from an external point to a circle are equal.


For circle $$C_1 , PB = PD$$ __(i)


and the circle $$C_2 , PA = PC$$ __(ii)


Subtracting (i) from (ii), we have 


$$PA - PB = PC - PD$$


$$\Rightarrow AB = CD$$


Hence, proved.


1795713_1603132_ans_9219878cb88d47619ad8a39e45492ee8.png

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