Proving angle congruence using circles and triangles

Question

The above picture is of an ellipse with foci G and F. AC and BD are tangent to the ellipse at the two ends of the major axis and CD is tangent to the ellipse at E. There are two circles, circle I and circle J. $\angle CFD=\angle DGC=90^{\circ}$.

I was asked to prove that $\angle ACF=\angle DCG$ and $\angle CDF=\angle BDG$.

Here is what I know: $\angle EDH=\angle EGH$ because they intersect the same arc. $\angle ECH=\angle EFH$ because they intersect the same arc. $\angle DBG$ and $\angle CAF$ are right angles because they are tangents of a point at the end of an axis.
If I can prove that $\angle CFA=\angle GDC$, I can show that $\angle ACF=\angle DCG$ because they would be congruent parts of similar triangles. If I can show that $\angle EFH=\angle FCA$ I can show that $ACF=DCG$. But I am stuck at this point. Any ideas of how to proceed?

Answer 1

$DGFC$ is cyclic. So $$\angle CDF = \angle CGF = \angle BDG$$

Also $$\angle DCG = \angle DFG = \angle ACF$$

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Thus it follows that $HG$, $HF$, $HE$ are angle bisectors and H is incenter of $\triangle EGF$.

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Explanation :

• $D, G, F, C$ lie on a circle since $\angle DGC=90=\angle DFC$ subtended by chord $DC$. Thus $DGFC$ is cyclic. $DC$ is diameter.
• $\angle CDF=\angle CGF$ - angles subtended by chord $CF$ of semicircle $DGFC$.
• $\angle BDG + \angle BGD=90$ Also $\angle BGD + \angle CGF=90$ $\Rightarrow \angle BDG=\angle CGF$
• $\angle CDF=\angle EDH=\angle EGH=\angle CGF \Rightarrow HG$ is angle bisector of $\angle EGF$.

Can you complete this now?