Any Solution?

parmenides51 wrote: In the cyclic quadrilateral $ABCD$, the sides $AB, DC$ meet at $Q$, the sides $AD,BC$ meet at $P, M$ is the midpoint of $BD$, If $\angle APQ=90^o$, prove that $PM$ is perpendicular to $AB$. A neat but hard problem

let $E$ the feet of the perpendicular from $P$ to $AB$. we will prove that $PE$ bisects $DB$. let $M$ be intersection of $PE$ and $DB$. and let $\angle APE =\alpha $ and $\angle EPB= \beta$. now note that $\angle PQA=\angle APE=\alpha=90-\angle PAB$. and $\beta=90-\angle PBA=\angle PQD$. by sines law $\frac {AQ}{DQ}=\frac{\sin \angle ADQ}{\sin \angle PCD}=\frac{PC}{PD}$. by that famous lemma : $\frac{AQ}{DQ} \times \frac{\sin \alpha}{\sin \beta} = \frac{PA}{PD}$. replacing $\frac{AQ}{DQ}$ with $\frac{PC}{PD}$ we get $\frac{PA}{PC}=\frac{\sin \alpha}{\sin \beta}$. and now in triangle $DPB$: $\frac{DM}{MB}=\frac{DP}{PB} \times \frac{\sin \alpha}{\sin \beta}=\frac{DP}{PB} \times \frac{PA}{PC}=1 \implies DM=MB$ Done .

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Kamran011 wrote: I found that the foot of the altitude from $P$ to $CD$ , $AC$ and $BD$ concur , maybe we can proceed ? No. It's False. See Attachment.

Attachments:

You know what's missing here? A moving point's sollution . So here it is:

Note, that WLOG $M$ is the midpoint of $AC$ in the figure posted @above and we want to prove, that $PM\perp CD$. Sketch of this statement

Fix $PDQ$ in place and animate $A$ on $PD$ with degree 1. As $ABCD$ is concyclic, angle $|\angle ABC|$ is fixed. Hence map from $A$ to $C$ is projective (using point's at infinity and fixed rotation). Hence degree of $C$ is 1. Degree of midpoint of two moving points is at most sum of their degrees, hence degree of $M$ is 2. We want to prove, that points $P$, $M$ and point at infinity perpendicular to $DQ$ are colinear. These have degrees 0, 2 and 0, hence it sufficies to check 3 cases. Trivialy check for $A=P$, $A=D$ and $P$ is the midpoint of $DA$. $\square$

Let $E=PQ\cap (ABP)$ Construct point $G$ on $AB$ such that $PG\perp AB$ Let $F=PG\cap CE,M'=PG\cap BD$ Since $\angle PEA=\angle PBA=\angle PDQ\implies ADQE$ is cyclic thus $\angle ADE=\angle AQE\implies \Delta AEP\cong \Delta EQB\implies \angle DEP=\angle BEQ$ Otherwise $\angle BEQ=\angle PAQ=\angle GPQ\implies \angle DEP=\angle GPQ\implies F$ is the midpoint of $DE$ $\implies M'$ is the midpoint of $BD$ (since $M'F\parallel BE$) thus $M'=M\implies PM\perp AB$

Attachments:

Drop perpendicular line $DE$ from $D$ to $AB$. Join $PE$. In the cyclic quadrilaterals $DEQP$ and $ABCD$, we have $\angle PEB=\angle PDQ=\angle PBE$. Thus, $\triangle PBE$ is isosceles. Let $F$ be the midpoint of $BE$. Then both $PF$ and $MF$ are perpendicular to $AB$. Thus, $P, M, F$ are collinear, and $PM$ is perpendicular to $AB$. [asy][asy] size(8cm); pair A = dir(110); pair D = dir(210); pair B = dir(330); pair E = foot(D,A,B); pair F = (E+B)/2; pair P = extension(A,D,F,F+dir(90)*(A-B)); pair Q = extension(A,B,P,P+dir(90)*(A-D)); pair C = extension(D,Q,B,P); pair M = (B+D)/2; draw(A--P--Q--A--cycle); draw(P--B--D); draw(D--E--P,dashed); draw(P--F,dotted); draw(D--Q); draw(unitcircle); dot("$A$",A,dir(A)); dot("$B$",B,dir(B)); dot("$C$",C,dir(C)); dot("$D$",D,dir(D)); dot("$E$",E,dir(E)); dot("$F$",F,dir(F)); dot("$M$",M,dir(M)); dot("$P$",P,dir(P)); dot("$Q$",Q,dir(Q)); [/asy][/asy]