I have been tasked to prove the following:
$$\triangle ABC \cong\triangle EDC $$
Give that $C$ is midpoint of $\overline{BE}$, and angles $\angle B $ and $\angle E$ are right angles.
How would you approach in proving the congruence?
P.S Drawing is not accurate representation
Proving triangle congruence
euclidean-geometrygeometry
Related Solutions
Your statement follows from the following nice lemma:
Let $ABCD$ be an isosceles trapezoid with $(AD)\parallel(BC)$ and let $E$ and $F$ be arbitrary diametrically opposite points on its circumcircle. Further let $G=(AE)\cap(BF),H=(CF)\cap(DE),I=(AC)\cap(BD)$.
Then: $I$ is the midpoint of $GH$.
To my sorry I did not find a geometrical proof of this lemma, and can present only an algebraic one.
Without loss of generality we may assume that the circumcircle radius is $1$. In the cartesian coordinate system with origin at the circumcenter and the axis $x$ directed parallel to $AD$ the points of interest have the following coordinates: $$\begin{array}{crr} &x\quad&y\quad\\ A&\cos\phi_1&\sin\phi_1\\ B&\cos\phi_2&\sin\phi_2\\ C&-\cos\phi_2&\sin\phi_2\\ D&-\cos\phi_1&\sin\phi_1\\ E&\cos\phi_{\phantom0}&\sin\phi_{\phantom0}\\ F&-\cos\phi_{\phantom0}&-\sin\phi_{\phantom0}\\ \end{array}$$
Then solving the systems of linear equations: $$ \begin{cases} \displaystyle\frac{x_G-x_E}{x_A-x_E}=\frac{y_G-y_E}{y_A-y_E}\\ \displaystyle\frac{x_G-x_F}{x_B-x_F}=\frac{y_G-y_F}{y_B-y_F}\\ \end{cases},\quad \begin{cases} \displaystyle\frac{x_H-x_E}{x_D-x_E}=\frac{y_H-y_E}{y_D-y_E}\\ \displaystyle\frac{x_H-x_F}{x_C-x_F}=\frac{y_H-y_F}{y_C-y_F}\\ \end{cases} ,\quad \begin{cases} \displaystyle\frac{x_I-x_A}{x_C-x_A}=\frac{y_I-y_A}{y_C-y_A}\\ \displaystyle\frac{x_I-x_B}{x_D-x_B}=\frac{y_I-y_B}{y_D-y_B}\\ \end{cases}, $$ one finds after boring but straightforward algebra: $$\begin{array}{ccc} &x\quad&y\quad\\ G&\displaystyle\hphantom-\frac{\cos\frac{\phi_1+\phi_2}2+\sin\frac{\phi_1-\phi_2}2\sin\phi}{\cos\frac{\phi_1-\phi_2}2}& \displaystyle\frac{\sin\frac{\phi_1+\phi_2}2-\sin\frac{\phi_1-\phi_2}2\cos\phi}{\cos\frac{\phi_1-\phi_2}2}\\ H&\displaystyle-\frac{\cos\frac{\phi_1+\phi_2}2+\sin\frac{\phi_1-\phi_2}2\sin\phi}{\cos\frac{\phi_1-\phi_2}2}& \displaystyle\frac{\sin\frac{\phi_1+\phi_2}2+\sin\frac{\phi_1-\phi_2}2\cos\phi}{\cos\frac{\phi_1-\phi_2}2}\\ I& 0&\displaystyle\frac{\sin\frac{\phi_1+\phi_2}2}{\cos\frac{\phi_1-\phi_2}2}\\ \end{array}$$
so that the lemma is proved.
In what follows I refer to your first figure.
To apply the above lemma it is useful to reverse your statement and let the point $D$ be the intersection of the line $(AB)$ with the circle $(AB'C')$. Then we need to prove that the point $M$ is on the line $C'D$. The corresponding trapezoid is the quadrilateral $B'C'ED$ on your figure, whereas the points $A$ and $K$ are the required antipode points.
Your proof looks okay to me, provided the notion that halves of congruent angles are congruent is in play. (At the foundations level, it can be tricky to know what's allowed, since different authors can present results in different orders.)
Note that your comment to MathMan suggests a way forward that's "safer" in this regard: Once you have $\triangle ADC\cong\triangle A'D'C'$, you can say $\angle A\cong\angle A'$ so that $\triangle ABC\cong\triangle A'B'C'$ by SAS. No subtle angle-algebra required.
Best Answer
Hint: you know the length of a side and two angles in both triangles.
$\angle B = \angle E$
$|BC|=|CE|$
Then you'll just need to argue that: $\angle ACB = \angle ECD$