Similar Right Triangles on Sides of Triangle and Treating BC as a Pascal Line to Obtain a Cyclic Hexagon Proposition. Let $ABC$ be an acute triangle. Exterior points $B’$ and $C’$ are such that $\triangle AB’B$ and $\triangle AC’C$ are similar with right angles at $B’$ and $C’$, respectively. If $M$ is the midpoint of $BC$ and $C’M \cap AB = D$, then $AB’DC’$ is a cyclic quadrilateral. Furthermore, if $B’M \cap AC = E$ and $B'B \cap C'C = K$, then the points $A$, $B'$, $C'$, $D$, $E$ and $K$ are concyclic.

How can I prove this? (I’m not sure what I can do to let the condition that M is a midpoint come into play.)
I need this to finish off a problem:
(Serbia 2018, Drzavno VI). Let $ABC$ be an acute triangle and let $AX$ and $AY$ be rays, such that the angles $\angle XAB$ and $\angle YAC$ have no common interior point with $\bigtriangleup ABC$ and $\angle XAB = \angle YAC < 90°$. Let $B'$ and $C'$ be feet of the perpendiculars from $B$ and $C$ to $AX$ and $AY$, respectively. If $M$ is the midpoint of $BC$, prove that $\overline{MB'} = \overline{MC'}$.

I wanted to show that the triangles in red are similar by proving two of the interior angles equal. As a consequence, triangles $NC’C’_R$ and $MC’B’$ would be similar as well, thus both isosceles. I’ve done it for the blue one and I need the cyclicity to be true for the other. There is perhaps an easier way to have dealt with the problem but I didn’t want to stop since I’ve reached this point. The cyclicity also seems quite nice. Is it actually well-known?
 A: 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.
