5

$k_1$ is a circle with center $O_1$ and radius $r_1$. Similar for $k_2(O_2;r_2)$. $r_1 < r_2$.

$AB$ and $CD$ are tangent lines to $k_1$ and $k_2$.

Image of the geometric situation

Prove that $AP=DQ$.

celtschk
  • 43,384

1 Answers1

2

From the Power of a Point Theorem it follows that:

$AB^2=AQ \cdot AD$ and $DC^2 =AD \cdot DP$.

Now it is not hard to see that $AB=DC$. It follows that $AQ=DP$, that is $AP+PQ=PQ+DQ$. Hence $AP=DQ$.

Nicky Hekster
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