TITLE: Similar Right Triangles on Sides of Triangle and Treating BC as a Pascal Line to Obtain a Cyclic Hexagon
QUESTION [9 upvotes]: 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?

REPLY [3 votes]: We will deploy the following lemma.
Lemma. Let $ABCD$ be an isosceles trapezoid ($AB \parallel CD$). Points $E$ and $F$ on ($ABCD$) are antipodes. If $AF \cap DE = G$, $BF \cap CE = H$ and $AC \cap BD = I$, then $I$ is the midpoint of the segment $GH$.
Proof. Let $O$ be the circumcenter of $(ABCD)$ and $JK$ the diameter perpendicular to the parallel sides of $ABCD$. Let $AK \cap DJ = L$ and $BK \cap CJ = M$. Furthrmore, let $HM \cap (ABCD) = P$. Note that $G-I-H$ and $L-I-M$ are collinear sets of points as an immediate consequence of Pascal's Theorem. By symmetry, $JLKM$ is a kite and $LM \perp JK$, so $LM \parallel CD$. Since $JK$ bisects $CD$, it bisects $LM$ as well, i.e. $IL = IM$. We claim that $\triangle ILG \cong \triangle IMH$.

We already have $IL = IM$ and $\angle GIL = \angle HIM$, it suffices to have $GL \parallel HM$. Since $E$ and $F$ are antipodes, $EJ = FK$, so $\angle GDL \equiv \angle EDJ = \angle FAK \equiv \angle GAL$, i.e. $ADLG$ is cyclic. Similarly, $BCHM$ is cyclic. Then,
$\begin{align*}
\quad \angle GLM + \angle PML
&= (\angle GLJ + \angle JLM) + (\angle JML +\angle PMJ) \\
&= \angle DAF + \angle JDC + \angle JCD + \angle  HMC \\
&= (\angle DBF + \angle FBC) + \angle JDC + \angle JCD \\
&= \angle DBC + \angle JDC + \angle JCD  \\
&= \angle DJC + \angle JDC + \angle JCD \\
&= 180°.
\end{align*}$
Therefore, $\triangle ILG \cong \triangle IMH$ and $IG = IH$, i.e. $I$ is the midpoint of $GH$.
Back to the original cyclicity we wish to show, we define $B'$ and $C'$ as needed and let $AB \cap (AB'C') = D$, $AC \cap (AB'C') = E$. We claim that $M \in C'D$ and $M \in B'E$. Let $C'D \cap B'E = M'$. Since $\angle C'AE \equiv \angle C'AC = \angle B'AB \equiv \angle B'AD$, we have $C'E = B'D$, i.e. $B'C'ED$ is an isosceles trapezoid. Now, we apply the lemma above with the antipodes $A$ and $B'B \cap C'C = K$ to say that $M'$ is the midpoint of $BC$, i.e. $M' \equiv M$, as desired.