Prove $EB=EC$ and that $F,M,G,C$ are concyclic in the given figure

Question

Given is a quadrilateral $ABCD$ in which $\angle DAB=\angle CDA=90$. Point M is the midpoint of side $BC$ and circumscribed circles of triangles $\triangle ABM$ and $\triangle DCM$ meet at points $M$ and $E$. The line $EC$ intersects the circumscribed circle of $ABM$ second time in $F$, and lines $AF$ and $CD$ meet at $G$. Prove $EB=EC$ and that $F,M,G,C$ are concyclic.

I've only worked on $EB=EC$ as I assume that's a prerequisite to $F,M,G,C$ being proven to be concyclic. I think we might have to prove that $E$ has to be on line $AD$ as we know that since $\triangle EBC$ should be an isosceles triangle and $M$ is midpoint of $BC$ the angles $\angle EMC=\angle EMB=90$, but also we know that since $CDEM$ is an cyclic quadrilateral that $\angle EMC+\angle EDC=180$ and $\angle EDC=\angle CDA-\angle ADE$ so $\angle ADE=0$? Any help's appreciated, thanks!

Answer 1

Hint:

Prove that $\angle AFM= \angle DCM$ by considering their relationships with $\angle ABC$.

Answer 2

If you proved $EB = EC$ the second one follows easily from there (couldn't comment cause of low rep.). If $EB=EC$ notice that $E$ has to lie on $AD$ since we have $\angle EMC= \angle CDA=90$ (basically EC and EB will be the diameters of the circles).
To prove points F, M, C, G are concyclic you can simply assign few angles to see it. Lets say $\angle EAF = \alpha, \angle EFA = \beta, \angle FCM = \theta$. Since $\angle DEC = \alpha + \beta$ and $\angle CEM = 90 - \theta$ angle $\angle DCM = 90 - \alpha - \beta + \theta$. Notice that angle $\angle GFM = 90 - \theta + \alpha + \beta$. Hence, those points are concyclic.

Answer 3

We will make use of a couple of corollaries to the Inscribed Angle Theorem. One corollary says that a quadrilateral is concyclic iff opposite angles are supplementary. Another is Thales' Theorem, which says that if the diameter of a circle is a side of an inscribed triangle, that triangle is a right triangle whose hypotenuse is the given diameter. Conversely, the hypotenuse of a right triangle is the diameter of its circumcircle.

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Let $O_1$ be the circumcenter of $\triangle ABM$ and $O_2$ be the circumcenter of $\triangle MCD$.

Let $E$ be the intersection of the circumcircles of $\triangle ABM$ and $\triangle MCD$.

Since $A,B,M,E$ are concyclic, $$ \angle AEM+\angle ABM=\pi\tag1 $$ Since $C,M,E,D$ are concyclic, $$ \angle MED+\angle MCD=\pi\tag2 $$ Since $\angle ABM+\angle MCD=\pi$, $(1)$ and $(2)$ give $$ \angle AED=\angle AEM+\angle MED=\pi\tag3 $$ That is, $A,E,D$ are colinear.

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Since $\triangle EAB$ is a right triangle, $O_1$ is the midpoint of diameter $\overline{EB}$.
Since $\triangle EDC$ is a right triangle, $O_2$ is the midpoint of diameter $\overline{EC}$.

Since $\overline{O_1B}=\frac12\overline{EB}$ and $\overline{BM}=\frac12\overline{BC}$ and $\angle O_1BM=\angle EBC$, SAS similarity says $$ \overline{O_1M}=\frac12\overline{EC}\tag4 $$ Since $\overline{O_1M}=\overline{O_1B}=\frac12\overline{EB}$, $(4)$ implies $$ \overline{EB}=\overline{EC}\tag5 $$

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Since $A,B,M,F$ are concyclic, $\angle AFM+\angle ABM=\pi$. Furthermore, $\angle AFM+\angle MFG=\pi$. Therefore, $$ \angle ABM=\angle MFG\tag6 $$ Furthermore, $\angle MCG+\angle ABM=\pi$. Therefore, $$ \angle MCG+\angle MFG=\pi\tag7 $$ which says that $M,F,G,C$ are concyclic.