We claim that $PE$ is the perpendicular bisector of $QR$. Indeed, note that $\angle PAQ=180^{\circ}-\angle BAC=180^{\circ}-\angle BDC=\angle PDR$, so $\triangle PAQ$ is congruent to $\triangle RDP$, thus $PQ=RP$. Also note that $\angle EDR=180^{\circ}-\angle EDC=\angle EAP$, and $EA=EF$, so $\triangle EDR$ is congruent to $\triangle EAP$ so $ER=EP$, and similarly $EQ=EP$, so $EQ=ER$, so $PE$ is the perpendicular bisector as desired.

There's a typo in the second line, it should be $AQ=DP$. So now, clearly $\angle PAQ=\angle RDP$ and so we get $\triangle PDR \cong \triangle QAP$ and so $PQ=PR$. Now, $AE=DE$, $AQ=PD$ and simple angle chasing gives $\angle QAE=\angle PDE$ and so $\triangle QAE \cong \triangle PDE$. Hence, $QE=PE$. Similarly, $RE=PE$ and so $QE=RE$ and hence $PE$ is the perpendicular bisector of $QR$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */import graph; size(8cm); real labelscalefactor = 0.5; /* changes label-to-point distance */pen dps = linewidth(0.7) + fontsize(0); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -14.734298959169909, xmax = 9.975643706670338, ymin = -10.658835600108045, ymax = 9.098692396418224; /* image dimensions */pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */draw(circle((-2,1.5), 4.912026058562801), linewidth(1) + wrwrwr); draw((-6.138342719649561,4.146151835160538)--(-0.14798681472621578,-3.0495106507812526), linewidth(1) + wrwrwr); draw((-6.138342719649561,4.146151835160538)--(1.9836447450819517,4.373773572322802), linewidth(1) + wrwrwr); draw((-4.48,5.74)--(-0.14798681472621578,-3.0495106507812526), linewidth(1) + wrwrwr); draw((-11.976521582716769,-1.4649745964558232)--(-4.48,5.74), linewidth(1) + wrwrwr); draw((1.9836447450819517,4.373773572322802)--(-0.8135739093779577,-5.367379400704707), linewidth(1) + wrwrwr); draw((-11.976521582716769,-1.4649745964558232)--(-0.8135739093779577,-5.367379400704707), linewidth(1) + wrwrwr); draw((-3.7277499147975433,4.213709600327491)--(-5.775078189632267,-1.6427352198622729), linewidth(1) + wrwrwr); draw((-11.976521582716769,-1.4649745964558232)--(-5.775078189632267,-1.6427352198622729), linewidth(1) + wrwrwr); draw((-5.775078189632267,-1.6427352198622729)--(-0.8135739093779577,-5.367379400704707), linewidth(1) + wrwrwr); draw((-0.8135739093779577,-5.367379400704707)--(-3.7277499147975433,4.213709600327491), linewidth(1) + wrwrwr); draw((-3.7277499147975433,4.213709600327491)--(-11.976521582716769,-1.4649745964558232), linewidth(1) + wrwrwr); draw((-6.138342719649561,4.146151835160538)--(-5.775078189632267,-1.6427352198622729), linewidth(1) + wrwrwr); draw((-5.775078189632267,-1.6427352198622729)--(-0.14798681472621578,-3.0495106507812526), linewidth(1) + wrwrwr); /* dots and labels */dot((-4.48,5.74),dotstyle); label("$B$", (-4.803404280177151,6.10117825446503), NE * labelscalefactor); dot((1.9836447450819517,4.373773572322802),dotstyle); label("$C$", (2.0778455761327903,4.641519194035649), NE * labelscalefactor); dot((-0.14798681472621578,-3.0495106507812526),dotstyle); label("$D$", (-0.0073816530520405856,-3.6993897227036734), NE * labelscalefactor); dot((-6.138342719649561,4.146151835160538),dotstyle); label("$A$", (-6.758304807537931,4.406931130752356), NE * labelscalefactor); dot((-5.775078189632267,-1.6427352198622729),linewidth(4pt) + dotstyle); label("$E$", (-6.184867319512102,-2.0833386200854296), SW * labelscalefactor); dot((-3.7277499147975433,4.213709600327491),linewidth(4pt) + dotstyle); label("$P$", (-3.6304639637606844,4.432996471117166), NE * labelscalefactor); dot((-11.976521582716769,-1.4649745964558232),linewidth(4pt) + dotstyle); label("$Q$", (-12.675137070349889,-1.4317051109651702), W * labelscalefactor); dot((-0.8135739093779577,-5.367379400704707),linewidth(4pt) + dotstyle); label("$R$", (-1.049995267644456,-6.0192050151717975), SE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

Note that $\triangle PAQ \cong \triangle RDP, \triangle DRE \cong APE, \triangle AQE \cong \triangle DEP$, so $E$ is the circumcenter of isosceles $\triangle PQR$, so we're done.

Note that $E'=(ABCD)\cap (BPQ)$ is the rotation center sending $AQ$ to $DP$, so $E'A=E'D\implies E=E'$. Additionally $EQ=EP$. Similarly, we get $ER=EP$. But $PQ=PR$ since $\triangle APQ \cong \triangle DRP$, so $PE$ is the perpendicular bisector of $QR$, as desired.

I think $AQ=DP$ not $DR$...

@ above Yep it's $AQ=DP$, it's a typo.

Amir Hossein wrote: Let five points on a circle be labelled $A, B, C, D$, and $E$ in clockwise order. Assume $AE = DE$ and let $P$ be the intersection of $AC$ and $BD$. Let $Q$ be the point on the line through $A$ and $B$ such that $A$ is between $B$ and $Q$ and $AQ = DR$ Similarly, let $R$ be the point on the line through $C$ and $D$ such that $D$ is between $C$ and $R$ and $DR = AP$. Prove that $PE$ is perpendicular to $QR$. Note that $\triangle EAR \cong \triangle EDP$ and $\triangle EAP \cong \triangle EDQ$ by SAS congruency. Thus, $EP=EQ=ER$. Now observe that $\triangle PAR \cong \triangle QDP$ hence $PR=PQ$ and so $PE$ is the perpendicular bisector of $QR$.

MATH1945 wrote: I think $AQ=DP$ not $DR$... You are right, thanks. I just fixed the typo.

Just get the congruences $\triangle RDP \equiv \triangle PAQ$, $\triangle RED \equiv \triangle PEA$ and $\triangle DPE \equiv QAE$. This implies that $RPQ$ is an isosceles triangle, and $E$ is the circumcenter of it.

I'm using directed angles, Now, $\angle QAP=\angle BAC=\angle BDC=\angle PDR$. $AQ=PD$, $AP=DR$. So, $\triangle QAP$ and $\triangle PDR$ are congruent. so,$PQ=PR$ Lemma 1 : $E$ is the circumcenter of $\triangle PQR$. Proof: $\angle EAP=\angle EAC=\angle EDR$. $AE=DE$, $AP=DR$. So, $\triangle EAP$ and $\triangle EDR$ are congruent. so, $EP=ER$ . Similarly $EP=EQ$. so $E$ is the circumcenter of $\triangle PQR$ So, $EQ=ER$, and $PE$ is the perpendicular bisector of $QR$.

Attachments:

Claim 1: $\triangle EAQ\cong\triangle EDP$ and $\triangle EAP\cong\triangle EDR$ Proof: Note that in cyclic quadrilateral $EABD$, we have $\angle EDP=180^{\circ}-\angle EAB=\angle EAQ$. Thus, by SAS congruence, we have $\triangle EAQ\cong\triangle EDP$. Similarly, we can prove that $\triangle EAP\cong\triangle EDR$.$\Box$ As a result of the triangle congruences from Claim 1, we have $PE=EQ$ and $PE=ER$, respectively. Claim 2: $\triangle REP\cong \triangle QEP$ Proof: From the triangle congruences in Claim 1, we have $\angle RED=\angle AEP$ and $\angle DEP=\angle QEA$. Notice that $$\angle REP=\angle RED+\angle DEP=\angle PEA+\angle QEA=\angle PEQ.$$Also, we have $RE=EP=EQ$ so by SAS congruence, we have $\triangle REP\cong \triangle QEP$. $\Box$ By Claim 2, we have $RP=PQ$ so $\triangle PQR$ is isosceles and $\angle RPE=\angle EPQ.$ Let $PE$ intersect $RQ$ at point $X$. Note that $PX$ is an angle bisector to the base side in a isosceles triangle so $PX$ is also a perpendicular bisector to $RQ$, finishing the problem. $\blacksquare$

^^True but anyways here a solution probably similar to many above. Observe that $APEQ \equiv DREP \implies PQ=PR$,but $E$ is clearly the circumcenter of $\odot{PQR}$ thus the result.

[asy][asy] size(9cm); defaultpen(fontsize(10pt)); pair E = dir(90), A = dir(35), B = dir(5), C = dir(225), D = dir(145), P = extension(A,C,B,D), Q = A+dir(B--A)*abs(D-P), R = D+dir(C--D)*abs(B-P); dot("$A$", A, dir(35)); dot("$B$", B, dir(355)); dot("$C$", C, dir(225)); dot("$D$", D, dir(170)); dot("$E$", E, dir(105)); dot("$P$", P, dir(275)); dot("$Q$", Q, dir(70)); dot("$R$", R, dir(145)); draw(unitcircle); draw(E--D--C--P--D--R--E--P--B--A--P--Q--A--E--Q--R--P); [/asy][/asy] Claim 1: $QP = PR$. Proof. Since $DR = AP$, $DP = AQ$, and $$\angle QAP = 180^\circ - \angle BAC = 180^\circ - \angle BDC = \angle PDR,$$we have $\triangle APQ \cong \triangle DRP$ by SAS. The conclusion follows. $\square$ Claim 2: $\triangle ERD \cong \triangle EPA$ and $\triangle EDP \cong \triangle EAQ$. Proof. Note that $DE = AE$, $DR = AP$, and $$\angle EDR = 180^\circ - \angle EDC = \angle EAP,$$so $\triangle ERD \cong \triangle EPA$ by SAS. The other congruence can be proved analogously. $\square$ It follows from the triangle congruences that $ER = EP = EQ$, so $E$ is the circumcenter of $\triangle PQR$. The result is now obvious as $\triangle PQR$ is isosceles with $QP = PR$. $\blacksquare$

Let $PE$ meet $QR$ at $S$. Claim1 : $AEQ$ and $DEP$, $AEP$ and $DER$ are congruent. Proof : $AQ = PD$, $AE = DE$ and $\angle EAQ = \angle BDE = \angle PDE$ so $AEQ$ and $DEP$ are congruent. we will prove the others with exact same approach. Now we have $QE = PE = RE$ and $\angle QES = \angle 180 - \angle QEA - \angle AWP = \angle 180 - \angle PED - \angle DER = \angle RES$ so $ES$ is perpendicular bisector of $QR$. we're Done.

Notice $\triangle AEQ\cong\triangle DEP$ as $\measuredangle QAE=\measuredangle BAE=\measuredangle BDE.$ Similarly, $\triangle EDR\cong\triangle EAP$ so $QE=PE=RE.$ Also, $$\measuredangle QAP=\measuredangle BAC=\measuredangle BDC=\measuredangle PDR$$so $\triangle APQ\cong\triangle DRP$ and $PQ=PR.$ Hence, $\overline{PE}$ is the perpendicular bisector of $\overline{QR}.$ $\square$

Note that from cyclic $ABCD$, we have $\angle CDB=\angle CAB$, so $\angle PDR=\angle PAQ$. Additionally, due to the length conditions, $\triangle PDR$ is congruent to $\triangle QAP$. Therefore, $\triangle PQR$ is isoceles. Additionally, from cyclic $AEDC$, we have $\angle EAC=\angle EDR$, and since $ED=EA$ and $DR=AP$, we have $\triangle EDR$ congruent to $\triangle EAP$ so we also have $ER=EP$. However, using the other side $DEAB$, we can also conclude that $EQ=EP$, so we have $RE=EQ$. Since $\triangle PQR$ is isoceles and $RE=EQ$, we have that $PE$ is an altitude of $\triangle PQR$, so we are done.

Begin by noting that $\angle{BAC} = \angle{BDC}$ which means that $\angle{PAQ} = \angle{PDR}$. Since $AP = RD$ and $PD = QA$, it follows that $\triangle{APQ} \cong \triangle{DRP}$ by SAS. This means that $PQ = PR$, so it suffices to prove that $EQ = ER$. Let $\angle{CAE} = \alpha$ and $\angle{BAC} = \angle{BDC} = \beta$. It follows that $\angle{QAE} = 180-\alpha-\beta$ and $\angle{PDE} = \angle{CDE}-\angle{CDB} = 180-\angle{CAE}-\angle{CDB}=180-\alpha-\beta$. Since $\angle{QAE} = \angle{PDE}$ and $PD = AQ$ and $DE = AE$, we have that $\triangle{DPE} \cong \triangle{AQE}$ by SAS. It follows that $PE = EQ$. Note that $\angle{EDR} = 180-\angle{CDE} = 180-(180-\alpha) = \alpha = \angle{PAE}$. Since $PA = DR$ and $AE = DE$, we have that $\triangle{APE} \cong \triangle{DRE}$ by SAS. It follows that $ER = PE$. Therefore, we have $PE = EQ = ER$ and $PE \perp QR$ as desired.

This took 5 minutes Observe $\triangle APQ\cong\triangle DRP$. This implies $QP=PR$ $\triangle EDR\cong\triangle AEP$ Therefore by spiral similarity $\measuredangle EAD=\measuredangle EPR$ Similarly $\measuredangle EAD=\measuredangle EPQ$ This implies EP is angle bisector We are done

$$\triangle QAP \cong \triangle PDR$$give us $PQ=PR$ Now notice $\triangle EAP \cong \triangle EDR \implies EP=ER$ and similarly $EP=ER=EQ$ Which give us $PE \perp QR$

Claim I: $\triangle APE \cong \triangle DRE$ because $AP=DR, AE=DE$ and $\angle PAE=180^o-\angle CDE=\angle RDE$ $\implies PE=RE \implies \angle ERP=\angle EPR = \angle GPE$, so $\triangle QPF \cong \triangle RPF$ by SAS congruence criteria. So $PF\perp QR$. Claim II: $\triangle APQ \cong \triangle DRP$ by SAS congruence as $AP=DR$ and $AQ=PD$ also $\angle PAQ=180^o-\angle PAB=180^o-\angle PDC= \angle PDR \implies PQ=PR$, $\angle QPE = \angle RPE$ and PE=PE $\triangle PQE \cong \triangle PRE$ by SAS congruence criteria. We have $\triangle PQE \cong \triangle PRE$ and $\triangle QPF \cong \triangle RPF$ $\implies \triangle QFE \cong \triangle RFE$. So $EF\perp QR$. Since P and E are on perpendicular bisector of $QR$, $PE \perp QR$.

[See Attachment] Claim 1 : $PQ = PR$ Proof : in $ \triangle APQ $ and $ \triangle DRP $ . Since, $ PD = AQ $ , $DR = AP$ and $\angle PAQ=180^o-\angle PAB=180^o-\angle PDC= \angle PDR$ This proves our claim, implying $\triangle QAP \cong \triangle PDR \implies PQ = PR $ and since, $ \angle QPE = \angle RPE $ Let $ E' = QR \cap PE$ $\hspace{4mm}$ [Since our aim is to prove $PE' \perp QR$ ] Claim 2 : $ \triangle QPE' \cong \triangle RPE' $ Proof : Since, $ \angle QPE = \angle RPE , PQ = PR $ and PE' is common . This Proves our claim $ \implies \angle PE'Q = \angle PE'R = 90^\circ \implies \boxed {PE \perp QR} $ And we are Done!

Attachments:

first we can change the statement : PE is perpendicular to QR <=> PR^2-PQ^2=ER^2-EQ^2 call this equation 1 and it is easy to see that triangle QAP is congruent to triangle PDR so we have PQ=PR => equation 1 LHS=0=RHS? now we want to prove RE=EQ and also it is easy to see that triangle ERD is congruent to triangle EPA => ER =EP similarly,we have triangle QEA is congruent to triangle PED => QE=EP => WE HAVE ER=EP so the LHS of equation 1 is also 0 Q.E.D.