Prove that DE is the perpendicular bisector of GH

euclidean-geometrygeometry

While I was finding a solution for a question that was posted in StackExchange recently, I found this.

enter image description here

The circles with centres $D$ and $E$ in the picture are identical and the radius of these circles is $R$.
$GFIH$ is a square that is inscribed in the common area for these two circles.

I can see that $GH$ is bisected by the line joining the centres of the circles (segment $DE$). And also I proved it using geogebra.

I tried to prove it using geometry but I couldn't.

Here is my approach

enter image description here

I already know that the common chord $JK$ is perpendicular to the line joining the centres.

So I created a rhombus $DJEK$ the I found the lengths of the diagonals using $30^\circ-60^\circ-90^\circ$ triangle rule. But that didn't help me.

Could anyone help me to solve this problem (Prefered without trigonometry)

Thank you for attending my question !

Best Answer

There is a theorem that states that the perpendicular bisector of a chord passes through the center of a circle. So consider chord $GH$ for the circle with $E$ as the center. If we construct perpendicular bisector to $GH$, we know that it will go thru $E$. But because it's a square, this bisector will also be perpendicular bisector to $IF$. Applying the same theorem to the other circle, it must go thru the point $D$.

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