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It seems that if we have a construction like in the image above that the two red angles are equal. However, I do not know how to prove that statement since every promising attempt I took failed at the point where my prove should not work for at least some non cyclic quadrilaterals.

So does someone know how to show that those two angles are indeed equal?


The cool thing about cyclic quadrilaterals is that the properties they hold often work in reverse. For example, it's easy to prove that if $ABCD$ is cyclic, then e.g. $\angle ABC+\angle ADC=180^\circ$. However, it turns out the reverse is true as well: if $\angle ABC+\angle ADC=180^\circ$, then $ABCD$ is cyclic. This is immensely powerful when put to good use!

And now, the solution to the original problem. I claim that quadrilateral $GDFC$ is cyclic. To see this, remark that $$\angle GDF+\angle GCF=90^\circ+90^\circ=180^\circ,$$ so the opposite angles of the quadrilateral are supplementary, meaning the quadrilateral is cyclic as desired.

Now a simple angle chase yields $$\angle BAC=\angle BDC\equiv\angle GDC=\angle GFC$$ which is what we wanted.

  • $\begingroup$ This is such a beautiful solution. Thank you. $\endgroup$ – RedAngle Jan 2 '16 at 0:10

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