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Let $\Gamma_1$ and $\Gamma_2$ be two non overlapping circles with centers $O_1$ and $O_2$ respectively. From $O_1$, draw the two tangents to $\Gamma_2$ and let them intersect $\Gamma_1$ at points $A$ and $B$. Similarly, from $O_2$, draw the two tangents to $\Gamma_1$ and let them intersect $\Gamma_2$ at points $C$ and $D$.

Prove that $AB=CD$.

I've done some extensive angle chasing on this but have been unable to make any real progress. Can't decide whether $ABCD$ is supposed to be rectangle (as in my diagrams), a parallelogram or even a trapezium. There is a homothety taking one circle to the other but as far as I can see this doesn't help as we don't have a clearly defined center to this.

Any help/hints would be greatly appreciated.

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  • $\begingroup$ Have the radii the same length ? $\endgroup$ Jul 30, 2013 at 5:29
  • $\begingroup$ Not necessarily, otherwise it would be trivial! $\endgroup$
    – John Marty
    Jul 30, 2013 at 5:32

1 Answer 1

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$$\frac{|AM|}{|AP|} = \frac{|QT|}{|QP|} \qquad \implies \qquad |AB| = \frac{2\cdot \text{product of radii}}{\text{distance between centers}} = |CD|$$

Since the length of chord $AB$ is "symmetric" with respect to elements of $\bigcirc P$ and $\bigcirc Q$ ---that is, swapping the roles of the circles leaves the formula unchanged--- this length must match that of chord $CD$.

Chords formed by tangents

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  • $\begingroup$ Thanks, the diagram is great. How do you create them? Is there an easy way? $\endgroup$
    – John Marty
    Jul 30, 2013 at 6:30
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    $\begingroup$ @JohnMarty: I used a program called GeoGebra to create the diagram. $\endgroup$
    – Blue
    Jul 30, 2013 at 6:35
  • $\begingroup$ I actually answered a duplicate of this question here. While I like the argument by "formula symmetry" above, the other answer may be helpful, as well. $\endgroup$
    – Blue
    Dec 16, 2015 at 17:30

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