Let $B'$ be a point on $(ABCD)$ such that $BB'CA$ is an isosceles trapezoid, we have $B'C=AB=QC$ and $B'A=CB=PA$. Let $N$ be the second intersection of $(PAB')$ with $PQ$. We've $\angle{B'NP}=\angle{B'AD}=180^{\circ}-\angle{B'CD}$. So, $\angle{B'CD}=180^{\circ}-\angle{B'NP}=\angle{B'NQ}$. This means $N'$ also lies on $(B'CQ)$. Note that $CB'=CQ$ and $AB'=AP$. So $\angle{ANC}=\angle{B'NC}+\angle{B'NA}=\frac{\angle{B'NQ}+\angle{B'NP}}{2}=90^{\circ}$. Let $(ANC)$ intersects $PQ$ again at $M'$, note that $\angle{AM'C}=90^{\circ}$. It's enough to prove that $M'=M$, i.e. $M'$ is the midpoint of $PQ$. Note that we've $\angle{M'AC}=\angle{CNQ}=\angle{CB'Q}$. Reflect $A$ and $C$ over $M'$ to get $A'$ and $C'$, respectively. Also, reflect $B'$ over $AC$ to get $B_2$. We've $CB'=CQ,CA=CA'$ and $\angle{ACA'}=\angle{B'CQ}\Rightarrow \angle{B'CA}=\angle{QCA'}$. This gives $\triangle{B'CA}\equiv \triangle{QCA'}$. It follows that $\triangle{AB_2C}$ is the reflection of $\triangle{A'QC}$ over $CM'$. Similarly, $\triangle{CB_2A}$ is the reflection of $\triangle{C'PA}$ over $AM'$. Hence, $\angle{PAM'}=\angle{B_2AM'}=\angle{B_2AC}+\angle{CAM'}=\angle{B'AC}+\angle{CA'M'}=\angle{CA'Q}+\angle{CA'M'}=\angle{M'A'Q}$. So, $AP\parallel A'Q$. Since $M'$ is the midpoint of $AA'$, $M'$ is also the midpoint of $PQ$, done.

Solution: We flip $\Delta ABC$. Note that this preserve all the properties of the configuration. We use complex numbers to prove $\angle AMC=\pi/2$. Lowercase letters will denote the corresponding uppercase letter points in the configuration, let the cyclic quadrilateral $ABCD$ lie on the unit circle. Then since $d,c,q$ are collinear we have, \[\frac{d-c}{q-c}=\frac{\bar{d}-\bar{c}}{\bar{q}-\bar{c}}\implies \bar{q}=\frac{c-q+d}{cd}\]Again since $BC=CQ$, \[|b-c|=|c-q|\implies (c-b)(\frac{1}{c}-\frac{1}{b})=(\frac{1}{c}-\frac{c-q+d}{cd})(c-q)\]\[\implies q=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+c\]Similarly, \[p=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+a\]We can rotate the quadrilateral, so WLOG we can assume $b=1$. It can be seen that exactly one of them is $+$ and the other has $-$ sign. WLOG assume $p$ has $+$ sign. Therefore, \[m=\frac{p+q}{2}=\frac{\sqrt{d}(a-c)+c+a}{2}\]Let $\delta=\frac{2a-2m}{2c-2m}$, we wish to show $\delta+\bar{\delta}=0$. Now, \[\delta=\frac{2a+\sqrt{d}(c-a)-c-a}{2c+\sqrt{d}(c-a)-c-a}=\frac{(c-a)(\sqrt{d}-1)}{(c-a)(\sqrt{d}+1)}=\frac{\sqrt{d}-1}{\sqrt{d}+1}\]Hence, \[\bar{\delta}=\frac{\frac{1}{\sqrt{d}}-1}{\frac{1}{\sqrt{d}}+1}=\frac{1-\sqrt{d}}{1+\sqrt{d}}\]Thus, $\delta+\bar{\delta}=0$. And we are done. $\blacksquare$ Remark: I am a beginner in complex bashing, so...

YanYau wrote: Suppose $ABCD$ is a cyclic quadrilateral. Extend $DA$ and $DC$ to $P$ and $Q$ respectively such that $AP=BC$ and $CQ=AB$. Let $M$ be the midpoint of $PQ$. Show that $MA\perp MC$. From medians formula we need to prove $MA^2+MC^2=AC^2$.

ayan.nmath wrote: Solution: We flip $\Delta ABC$. Note that this preserve all the properties of the configuration. We use complex numbers to prove $\angle AMC=\pi/2$. Lowercase letters will denote the corresponding uppercase letter points in the configuration, let the cyclic quadrilateral $ABCD$ lie on the unit circle. Then since $d,c,q$ are collinear we have, \[\frac{d-c}{q-c}=\frac{\bar{d}-\bar{c}}{\bar{q}-\bar{c}}\implies \bar{q}=\frac{c-q+d}{cd}\]Again since $BC=CQ$, \[|b-c|=|c-q|\implies (c-b)(\frac{1}{c}-\frac{1}{b})=(\frac{1}{c}-\frac{c-q+d}{cd})(c-q)\]\[\implies q=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+c\]Similarly, \[p=\pm\frac{\sqrt{d}(b-a)}{\sqrt{b}}+a\]We can rotate the quadrilateral, so WLOG we can assume $b=1$. It can be seen that exactly one of them is $+$ and the other has $-$ sign. WLOG assume $p$ has $+$ sign. Therefore, \[m=\frac{p+q}{2}=\frac{\sqrt{d}(a-c)+c+a}{2}\]Let $\delta=\frac{2a-2m}{2c-2m}$, we wish to show $\delta+\bar{\delta}=0$. Now, \[\delta=\frac{2a+\sqrt{d}(c-a)-c-a}{2c+\sqrt{d}(c-a)-c-a}=\frac{(c-a)(\sqrt{d}-1)}{(c-a)(\sqrt{d}+1)}=\frac{\sqrt{d}-1}{\sqrt{d}+1}\]Hence, \[\bar{\delta}=\frac{\frac{1}{\sqrt{d}}-1}{\frac{1}{\sqrt{d}}+1}=\frac{1-\sqrt{d}}{1+\sqrt{d}}\]Thus, $\delta+\bar{\delta}=0$. And we are done. $\blacksquare$ Remark: I am a beginner in complex bashing, so... Good evening! We have BC = AP , not BC = CQ ! Although it is not a dammage for the whole idea!

@above Umm... Actually it is "Good night" for me . I wrote "We flip $\triangle ABC$" in the very first line of my proof, so I think it should be fine.

ayan.nmath wrote: @above Umm... Actually it is "Good night" for me . I wrote "We flip $\triangle ABC$" in the very first line of my proof, so I think it should be fine. Sorry, but I don't understand the term "flip" ! Im a beginner too, more than you!

Make a trapezoids $ABCP_{1}$ and $ABCQ_{1}$ such that $AP=BC=AP_{1}$ and $CQ=AB=CQ_{1}$.Let $CP_{1}\cap AQ_{1}=N$ then $180-\angle ADC=\angle ABC=\angle ANC=\angle P_{1}NQ_{1}$ wich gives $\angle AP_{1}N=\angle AP_{1}=\angle ADCC=\angle ADC=\angle P_{1}NA$ so $P_{1}A=AN=AP$ and similarly $NQ_{1}=NC=CQ$.Now $$\angle PDQ+\angle DPN+\angle DQN=\angle PNQ...(1)$$but $\angle PDQ=180-\angle ABC=180-\angle ANC=180-(\angle ANP+\angle PNQ+\angle CNQ)=180-(\angle DPN+\angle PNQ+\angle DQN )$ and replacing in $(1)$ gives $180 =2\angle PNQ$ so $\angle PNQ=90$ and so $MN=MP=MQ$ but we proved $PA=AN$ and $CN=CQ$ thus $MA$ bisects $\angle NMP$ and $MC$ bisects $\angle NMQ$ so $\angle AMC=\frac{\angle NMP+\angle NMQ}{2}=90$ as desired.$\blacksquare$

Let $K,N$ be the midpoints of $CA,CP$, then $\triangle ABC\sim \triangle MNK$ with ratio $2$, so $KM=\frac{AC}{2}=KA=KC$ $\implies MA\perp MC$.

Let $A'$ be the reflection of point $A$ through point $M$ $\implies$ $APA'Q$ is a parallelogram. Since $\angle APQ + \angle CQP=\angle ABC$ , it follows that that $\angle A'QC=\angle CQP+\angle A'QP=\angle CQP+\angle APQ=\angle ABC$.$\implies$ $\boxed{\angle A'QC=\angle ABC}$. And we also have that $A'Q=AP$ and $CQ=AB$ $\implies$ $\boxed{\triangle ABC= \triangle A'QC}$ $\implies$ $AC=A'C$ $\implies$ $CM\perp AM$

Here is a routine spiral similarity solution. Let $T = \odot(DAC)\cap \odot(DPQ)$ be the spiral center. Then, since $\triangle TAP\sim\triangle TCQ$, $\tfrac{TA}{TC} = \tfrac{AP}{CQ} = \tfrac{BC}{AB}$, which means that $BT$ passes through the midpoint $K$ of $AC$. Seeing this, we observe that $\triangle TAP\sim\triangle TKM$ or $$\frac{KM}{TK} = \frac{AP}{AT} = \frac{BC}{AT} = \frac{CK}{KT}$$or $KM=KC=KA$ as desired.

See my solution to this problem on my Youtube channel here: https://www.youtube.com/watch?v=5aTTapplHQQ

Let $B'$ be the reflection of $B$ about perpendicular bisector of $AC$. Let $A,C$ be the midpoints of $PS,QT$ respectively. So $\triangle B'PS\overset{+}{\sim}\triangle B'TQ$ hence $B',P,S,PT\cap QS$ are cyclic. The fact that $AM//QS$ and $MC//PT$ finishes.