Let $AB$ be a chord of the semicircle $O$ (not the diameter). $M$ is the midpoint of $AB$, and $D$ is a point lies on line $OM$ ($D$ is outside semicircle $O$). Line $l$ passes through $D$ and is parallel to $AB$. $P, Q$ are two points lie on $l$ and $PO$ meets semicircle $O$ at $C$.
If $\angle PCD=\angle DMC$, and $M$ is the orthocentre of $\triangle OPQ$. Prove that the intersection of $AQ$ and $PB$ lies on semicircle $O$.
The angle condition implies that $(CMD)$ is tangent to $OP$ and therefore $OM\cdot OD=OC^2$. Therefore $l$ is the polar of $M$ with respect to $O$. Now define $A'$ and $B'$ to be the second intersection of $PA$ and $PB$ with $O$, respectively. By Brocard $A'B'\cap AB$ must lie on the polar of $P$ with respect to $(O)$, and the only intersection of this polar with $AB$ is $M$, so $A',B',M$ are collinear. Since $M$ is the orthocenter of $OPQ$, by Brocard again $AB'\cap BA'=Q$, which implies $PB\cap AQ=B'$ is on $(O)$.
Suppose that $AP\cap MQ=X,BQ\cap MP=Y,AP\cap \odot(O)=T,BQ\cap\odot (O)=T'$,Let $PM\cap OQ=E,QM\cap OP=F$
Note that $OC^{2}=OM\cdot OD=OA^{2}=OB^{2}$
Consider that $\angle BMD=90^{\circ}$
Hence $\angle OBD=\angle OAD=90{\circ}$
So $PQ$ is the polar of $M$ respect to $\odot(O)$
Suppose $QO\cap \odot(O)=R$
Since $OC^{2}=OE\cdot OQ$
Hence $-1=(Q,E;C,R)$
So $MP$ is the polar of $Q$ respect to $\odot(O)$
Similarly $MQ$ polar of $P$ respect to $\odot(O)$
Which means that $-1=(Q,Y;T',B)=(P,X;T,A)$
Thus $T=T'$
Which means that the intersection of $AQ$ and $PB$ lies on semicircle $O$
Let semicircle $O$ be on the unit circle, so that
$$o=0$$$$|a|=|b|=|c|=1$$$$m=\frac{a+b}2$$Since $\angle DCP = \angle DMC$ as directed angles, we have
$$\frac{\left(\frac{d-c}{p-c}\right)}{\left(\frac{d-m}{c-m}\right)} \in \mathbb{R}$$Now $\frac{p-c}c, \frac{d-m}m \in \mathbb{R}$. Thus we write
$$\frac{(c-d)(c-m)}{cm} \in \mathbb{R}$$Since $\overline{d} = \frac{d}{ab}$, this becomes
$$\frac{(c-d)(2c-a-b)}{c(a+b)} = \frac{(ab-cd)(2ab-ac-bc)}{abc(a+b)}$$$$ab(c-d)(2c-a-b) = (ab-cd)(2ab-ac-bc)$$$$d = \frac{ab(2ab-ac-bc) - abc(2c-a-b)}{c(2ab-ac-bc) - ab(2c-a-b)} = \frac{2ab(ab-c^2)}{(a+b)(ab-c^2)} = \frac{2ab}{a+b}$$Now $P$ lies on line $OC$, so we have
$$\overline{p} = \frac{p}{c^2}$$But line $PD$ is also parallel to line $AB$, so
$$\frac{d-p}{a-b} \in \mathbb{R}$$$$\frac{2ab-ap-bp}{(a-b)(a+b)} = \frac{ab(ap+bp-2c^2)}{c^2(a-b)(a+b)}$$$$2abc^2-ac^2p-bc^2p = a^2bp+ab^2p - 2abc^2$$$$p = \frac{4abc^2}{(a+b)(ab+c^2)}$$Now we also know, since $M$ is the orthocenter of $\triangle OPQ$, that $QM \perp CO$. Thus
$$\frac{q-m}{c-o} \in i\mathbb{R}$$$$\frac{2q-a-b}{2c} = \frac{c(a+b-2ab\overline{q})}{2ab}$$$$\overline{q} = \frac{(a+b)(ab+c^2) - 2abq}{2abc^2}$$Also, $Q$ lies on line $\ell$, so
$$\frac{d-q}{a-b} \in \mathbb{R}$$$$\frac{2ab-aq-bq}{(a-b)(a+b)} = \frac{ab(a\overline{q}+b\overline{q}-2)}{(a-b)(a+b)}$$$$\overline{q} = \frac{4ab-aq-bq}{ab(a+b)}$$Setting these equal gives
$$\frac{(a+b)(ab+c^2) - 2abq}{2abc^2} = \frac{4ab-aq-bq}{ab(a+b)}$$$$(a+b)^2(ab+c^2) - 2abq(a+b) = 8abc^2 - 2c^2q(a+b)$$$$q = \frac{(a+b)^2(ab+c^2) - 8abc^2}{2(a+b)(ab-c^2)}$$Now let $R$ be the second intersection of line $PB$ with the unit circle. Then
$$r = \frac{b-p}{b\overline{p}-1} = \frac{b-\frac{4abc^2}{(a+b)(ab+c^2)}}{\frac{4ab^2}{(a+b)(ab+c^2)}-1} = \frac{b(a^2b+ab^2+bc^2-3ac^2)}{3ab^2-a^2b-ac^2-bc^2}$$It remains to show that $A,Q,R$ are collinear. Indeed, we find the vectors
$$r-a = \frac{(a-b)^2(ab+c^2)}{3ab^2-a^2b-ac^2-bc^2}$$and
$$q-a = \frac{(a-b)(3ac^2-a^2b-ab^2-bc^2)}{2(a+b)(ab-c^2)}$$Thus
$$\frac{q-a}{r-a} = \frac{(3ac^2-a^2b-ab^2-bc^2)(3ab^2-a^2b-ac^2-bc^2)}{2(a+b)(a-b)(ab+c^2)(ab-c^2)}$$which is real. $\blacksquare$
Let $AB$ be a chord of the semicircle $O$ (not the diameter). $M$ is the midpoint of $AB$, and $D$ is a point lies on line $OM$ ($D$ is outside semicircle $O$). Line $l$ passes through $D$ and is parallel to $AB$. $P, Q$ are two points lie on $l$ and $PO$ meets semicircle $O$ at $C$. If $\angle PCD=\angle DMC$, and $M$ is the orthocentre of $\triangle OPQ$. Prove that the intersection of $AQ$ and $PB$ lies on semicircle $O$.
