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For a week I've been struggling with this problem. I tried to solve it with angles, triangle congruences, Pitagora's theorem.

I'm supposed to prove that $EH=\frac{R}{2}\sqrt{10-2\sqrt{5}}$, where $R$ is the common radius of both circles with the centers $A$ and $B$. The big circle with the center also in $B$ has the radius $R^{'}=EC$. This problem shows that $EH$ is a side of a regular pentagon inside the circle with center in $A$. Any hint would be very much appreciated.


Hint: Do you know the intersecting secants theorem? Apply it to the circle $ACD$ for secants $HC$ and $HB$ to find the length of $AH$, and then use the Pythagorean theorem to find $EH$.

  • $\begingroup$ Thanks for the tip. I managed to solve it. $\endgroup$ – O.Alexandra Apr 23 '18 at 19:54

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