$k_1$ 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 $AP=DQ$.

  • $\begingroup$ is O, O1, O2 a line? $\endgroup$ – JMP Apr 28 '15 at 20:15
  • $\begingroup$ Yes, because O is the intersection of the 2 tangents of the circles $\endgroup$ – imranfat Apr 28 '15 at 20:19
  • $\begingroup$ Hint: Calculate the "power" of point $D$ with respect to $k_1$, and the "power" of point $A$ with respect to $k_2$. $\endgroup$ – Blue Apr 28 '15 at 20:31
  • $\begingroup$ if you draw the 'inner tangential' from k1 to k2 (it passes close to AD) and say it intersects k1 at X and k2 at Y, does AX = YD? $\endgroup$ – JMP Apr 28 '15 at 20:34

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 $AP=DQ$.

  • 1
    $\begingroup$ Goed gedaan! +1 $\endgroup$ – imranfat Apr 28 '15 at 21:05
  • $\begingroup$ @imranfat Bedankt! $\endgroup$ – Nicky Hekster Apr 28 '15 at 21:24

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