Question

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

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.Maths

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