From the given figure, prove that $AB=DC$

Question

In the given figure, $AG=FG$, $EG=CG$ and $DC\perp CE$. Prove that $AB=DC$.

My Attempt $$1. \measuredangle GAF=\measuredangle GFA$$ $$2. \measuredangle GEC=\measuredangle GCE$$

I could not get further from here.

Answer 1

Hint: triangles $\triangle AGF$ and $\triangle EGC$ are isosceles, therefore:

$$ \begin{align} \angle GAF = (\pi - \angle AGF)/2 \\ \angle GCB = (\pi - \angle CGE)/2 \\ \end{align} $$

Adding the two equalities, and noting that $\angle AGF + \angle CGE = \pi\,$:

$$ \angle GAF + \angle GCB = (2 \pi - \pi)/2 = \pi /2 $$

Then $\angle ABC = \pi - \angle GAF - \angle GCB = \pi - \pi / 2 = \pi/2$ so $\triangle ABC$ is a right triangle, and the center of the circle is the midpoint of the hypotenuse $AC\,$.

Answer 2

You have $$\angle AGF = 2 \angle GCE = 2(90^\circ - \angle ACD)$$, and $$2\angle FAG = 180^\circ - \angle AGF = 180^\circ - 2(90^\circ - \angle ACD) = 2\angle ACD$$, then $$\angle FAG = \angle ACD.$$

So we have $AB \parallel CD$, then $AB \perp BC$, or $AC$ is diameter of the circle. From here you can see that ABCD is a quadrilateral.