I have the following problem: Two circles (middle points $A,C$) have two intersections: $E,F$. Now we draw a common tangent on the circles and we'll get $G,H$. Let $I:=GH \cap EF$. Now I have to proof $GI=IH$.

I made this drawing:

enter image description here

and I tried to find some common angles or sides to use congruences, but I wasn't able to find some, because I have differnt circles with different radius'.

Perhaps someone can give me a hint how to move on? Thanks in advance.

  • 1
    $\begingroup$ If you want an "analytical" approach, I would take a gander at this mathworld.wolfram.com/Circle-CircleIntersection.html $\endgroup$
    – user39280
    Jun 30, 2013 at 21:41
  • $\begingroup$ Thanks for the link, but unfortunately I'm looking for a proof with triangles, congruences and so on. $\endgroup$
    – ulead86
    Jun 30, 2013 at 21:45

1 Answer 1


As $I$ is on the power line (also called radical axis) of the circles $(A)$ and $(C)$ its power with respect to each of the circles is the same. Thus, $IG^2 = IH^2\implies IG=IH$

EDIT: Without knowledge of the properties of radical axis, we can do the following. Show that triangles $\Delta IGE$ and $\Delta IFG$ are similar using the fact that $\angle IGE = \angle IFG$. Therefore, $\frac{IG}{IF}=\frac{IE}{IG}$ or equivalently $IG^2 =IE\cdot IF$. Similarly, we can show $IH^2=IE\cdot IF$ and the problem is solved.

  • $\begingroup$ I think the OP's question is equivalent to proving the common secant IF is the radical axis of $\,A\,,\,C\,$ ... $\endgroup$
    – DonAntonio
    Jun 30, 2013 at 21:58
  • $\begingroup$ excellent! I got it. $\endgroup$
    – ulead86
    Jul 1, 2013 at 6:07

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