First we know that $\angle{APE}=2\angle{ABE}=90^o$ and $\angle{AQE}=2\angle{ADE}=90^o$. Since $AE$ is the common chord of the two circumcircles, thus if we let $H=PQ \cap AE$, then we shall have $\angle{QHE}=\angle{PHE}=90^o$, hence $PE^2=EH \cdot EA=QE^2 \Longrightarrow PE=QE$. Therefore we get that $AP=PE=EQ=QA$. So APEQ is a rhombus . Remember that we already have $\angle{APE}=90^o$, so $APEQ$ is a square.

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My solution for this question, I remember: $\angle APE = 2\angle ABE = 90$ $\angle AQE = 2\angle ADE = 90$ $\triangle APE$ is isosceles as $AP=PE$ equal radii. $\implies \angle PEA=\angle PAE=45$ $\triangle AQE$ is isosceles as $QE=QA$ equal radii. $\implies \angle QEA=\angle QAE=45$ $\implies \angle QAP = \angle QEP = 90$ 4 right angles define a rectangle but we know $AP=PE$ A rectangle with a pair of adjacent sides equal is a square $QED$

Let X the orthogonal projection of P on AB, and let Y the orthogonal projection of Q on AD, We now that angle{APE}=angle{PAQ}=90^o , we now that AY=AD/2=AX, if angle{QAY}=T then angle{BAQ}=angle{XAQ}=90^o-T then angle{XAP}=T then triangle APX ~= triangle AQY then AP=AQ, => APEQ is a square

Hi my dears. I want to solve this easy and nice problem by inversion. I can consider that there not exists square$ABCD$, but there only exists isosceles right angle$ABD$. Ok , lets start to invert the shape. Suppose $B,P,E,Q,D$ , changes to $B',P',E',Q',D'$, respectively. ................................................................................................................................................................................ . Now the problem changes to this one: $AB'D'$ is a right angle triangle, $E'$, is a point on the circumcircle of$AB'D'$. $P'$ , is the respect of $A$ to line $B'E'$. and $Q'$ is the respect of $A$ to line $E'D'$. prove $P',E',Q'$ are collinear and $AP'Q'$ is an isocels right angle triangle. ……………………………………………………………………………………………………………………. ok, it is clear that $AE'P'$ is an isosceles triangle, and because of ,$\angle AE'B'=45$, then:$\angle AE'P'=90$ and $AE'P'$ is isosceles right angle triangle. With a similar way we can conclude that,$AE'Q'$ is too. Then every thing is ok, because: $AB'P'$ and $AD'Q'$ are equal.and $AP'=AQ'$ $\angle AE'P'=\angle AE'Q'=90$. Then $P'Q'E'$ are collinear. $\angle P'AE'=E'AQ'=45$, then $\angle P'AQ'=90$. And it is done.

I immediately tried coordinate geometry, takes a few minutes, but it works! Set A(0,1); B(0,0); C(1,0); D(1,1) and E(e,e). PQ must be the perpendicular passing through center of line AE, let this point be M. M will have coordinates (e/2, [e +1] /2). Thus PQ will then have the equation y = (e/1-e)x + [2e²-1] / [2(e-1)] P must have a y-coordinate of ½ ... circumcenter of ABE. Q must have a x-coordinate of ½ ... circumcenter of AED. By substitution this gives the coordinates of P( x-½, ½ ) , and Q( ½ , x + ½ ). Applying the distance formula AP = PE = EQ = QA = sqrt[x² - x + ½ ] Looking at the gradients of AP,PE, EQ, QA These all form angles of 90° Thus APEQ is a square Eulerian geometry solution looks a lot better.

An equivalent enunciation. Let $ABCD$ be a square and let $E$ be a point which belongs to the diagonal $[BD]$. Denote the line $d$ for which $E\in d$ , $d\perp AE$ and the intersections $R\in CB\cap d$, $S\in CD\perp d$. Prove that the points $A$, $E$ and the middlepoints of the segments $AR$, $AS$ are the vertices of a square. Remark. Thus, the our nice problem becomes very simply !

shobber wrote: hence $PE^2=EH \cdot EA=QE^2 \Longrightarrow PE=QE$. I think, that $PE^2=EH^2+PH^2=EH^2+EA^2$ and $QE^2=EH^2+QH^2=EH^2+EA^2$

Dear Mathlinkers, 1. ABD being A-right angle isoceles, PAQ is also A-right angle isosceles 2. PQ is the perpendicular bisector of AE 3. and we are done… Sincerely Jean-Louis

Call $R_1$ the circumradius of $(AEB)$ and $R_2$ the circumradius of $(AED)$. We know that $[AEB] = \frac{AE \cdot BE \cdot AB}{4R_1}$ and $[AED] = \frac{AE \cdot ED \cdot AD}{4R_2}$. Therefore, we have that: $$\frac{[AEB]}{[AED]} = \frac{\frac{AE \cdot BE \cdot AB}{4R_1}}{ \frac{AE \cdot ED \cdot AD}{4R_2}} = \frac{\frac{BE}{R_1}}{\frac{ED}{R_2}}$$We can also write: $$\frac{[AEB]}{[AED]} = \frac{BE}{ED}$$Hence, $$\frac{BE}{ED} = \frac{\frac{BE}{R_1}}{\frac{ED}{R_2}} \implies R_1=R_2$$Therefore, $AP=PE=EQ=AQ$. However, also note that the radical axis of $(AED)$ and $(AEB)$ is $AE$, so $AE \perp PQ$. Therefore, $APEQ$ is a square. $\blacksquare$

Since, $P,Q$ lie on perpendicular bisector of $AE$, thus $APEQ$ is a kite. Since, $\angle ADE$ $=$ $45^{\circ}$ $=\tfrac{1}{2}$ $\angle AQE$ $\implies$ $\angle AQE$ $=$ $90^{\circ}$. Similarly, $\angle APE$ $=$ $90^{\circ}$ $\implies$ $APEQ$ is square

Note that $\angle ADB=\angle ABD=45$ Then: $\angle APE=90$ In $\triangle AEB$, $Q$ is the circuncenter $AQ=QB=QE$ Then: $\angle AQE=90$ But $AQ=AE$ and $AP=PE$ $\angle PAE=\angle PEA=45$ and $\angle QAE=\angle QEA=45$ $\angle PAQ=\angle PEQ=90$ but $AQ=QE$ and $AP=AE$ $APEQ$ is square. Finish