I'll restate the problem with different notations. Problem: $ ABCDE$ is a pentagon and $X,Y,Z,W,P,Q$ are the orthogonal projections of $ E$ onto $ AB,$ $CD,$ $AC,$ $BD,$ $AD,$ $BC.$ Then $EX \cdot EY = EZ \cdot EW = EP \cdot EQ$ $\Longleftrightarrow$ $ ABCDE$ is cyclic. We assume that $ ABCD$ is cyclic. If $ EX \cdot EY = EP \cdot EQ$ $ \Longrightarrow$ $ \triangle EXP \sim \triangle EQY$ since $ \angle XEP = \angle QEY.$ Then $ E$ is the center of the spiral similarity that takes $ \triangle EXP$ into $ \triangle EQY$ $\Longrightarrow$ circles $\odot (EXP)$ and $\odot (EQY)$ meet on $QY$ and angle chase yields $ \angle EBA = \angle ADE$ $ \Longrightarrow$ $ E \in \odot(ABCD).$ It remains to prove $ EZ \cdot EW = EP \cdot EQ,$ which folllows from the fact that $ Y,W, Q$ are collinear on the Simson line with pole $ E$ WRT $ \triangle BCD$ and $Y,P,Z$ are collinear on the Simson line with pole $ E$ WRT $ \triangle ACD.$ Then $\odot (YWP)$ and $ \odot (YZQ)$ meet at $ E$ $\Longrightarrow$ $ E$ is the center of the spiral similarity taking $ \triangle EZP$ into $\triangle EQW,$ hence $ EZ \cdot EW = EP \cdot EQ.$

Besides this natural synthetic solution one could (not easily!) find the following one, which uses techniques of algebraic geometry. It involves some well-known properties of conics as well as some elements of complex geometry. I will write down the solution in different posts, for a better understanding.

At first I will reformulate the problem: Prove that pentagon $ A_0A_1A_2A_3A_4$ is cyclic iff the product $ \mathrm{dist}\(A_4,A_iA_j)\cdot\mathrm{dist}(A_4,A_kA_l)$ is the same for all choices of indices $ \{i,j,k,l\}=\{0,1,2,3\}.$

Solution. In what follows, we denote by lowercases the complex coordinates of points denoted by uppercases. The equation of the oriented line $ A_jA_k$ is given by $ f_{jk}(z,\bar{z})=\left|\begin{array}{ccc} a_j & \bar a_j & 1 \\ a_k & \bar a_k & 1 \\ z & \bar z & 1 \\ \end{array} \right|=0$ or $ f_{jk}=(\bar a_j-\bar a_k)z-(a_j-a_k)\bar z+a_j\bar a_k-\bar a_ja_k=0.$ The distance from $ A_4$ to $ A_jA_k$ is $ -\frac{f_{jk}(a_4,\bar a_4)}{2\mathrm{i}|a_j-a_k|}.$ Next, it is not hard to verify that $ f_{jk}((1-t)z+tz',(1-t)\bar{z}+t\bar z'=(1-t)f_{jk}(z,\bar z)+tf_{jk}(z',\bar z'), \forall t\in\mathbb{R}$ and $ z,z'\in\mathbb{C}$.

The condition in the statement is equivalent to the fact that $ A_4$ lie on the following 2 conics passing through $ A_0,A_1,A_2,A_3$: $ |a_0-a_3|\cdot|a_1-a_2|f_{01}f_{23}-|a_0-a_1|\cdot|a_2-a_3|f_{03}f_{12}=0\ \ (1)$ and $ |a_0-a_2|\cdot|a_1-a_3|f_{01}f_{23}-|a_0-a_1|\cdot|a_2-a_3|f_{02}f_{13}=0\ \ (2)$ This is possible iff the 2 conics coincide. Both of the conics belong to the pencil of conics determined by $ A_0,A_1,A_2,A_3.$ It is renowned that any two distinct conics of this pencil form a basis over $ \mathbb{R}$ for the pencil; in particular, any two of the three (degenerate) conics: $ f_{01}f_{23}=0,f_{02}f_{13}=0,f_{03}f_{12}=0.$ These three are therefore linearly dependent over $ \mathbb{R}$. Moreover, they satisfy, as we will show, the following Ptolemy-like relation: $ f_{01}f_{23}-f_{02}f_{13}+f_{03}f_{12}=0\ \ (\star)$

Assume $ (\star)$ for the time being to rewrite (2) as: $ (|a_0-a_2|\cdot|a_1-a_3|-|a_0-a_1|\cdot|a_2-a_3|)f_{01}f_{23}-|a_0-a_1|\cdot|a_2-a_3|f_{03}f_{12}=0\ \ (2')$ Now the two conics coincide iff: $ |a_0-a_1|\cdot|a_2-a_3|-|a_0-a_2|\cdot|a_1-a_3|+|a_0-a_3|\cdot|a_1-a_2|=0.$ By Ptolemy Theorem the points $ A_0,A_1,A_2,A_3$ are concyclic.

It only remains to establish $ (\star)$ for the proof to be complete. Notice that $ f=f_{01}f_{23}+f_{03}f_{12}$ vanishes at any point $ z$ of the conic $ f_{02}f_{13}=0$: for example, if $ z$ lies on the line $ f_{02}=0$ then $ z=(1-t)a_0+ta_2,$ for some $ t\in\mathbb{R}$ so that $ f(z,\bar{z})=t(1-t)(f_{01}(a_2,\bar a_2)f_{23}(a_0,\bar a_0)+f_{03}(a_2,\bar a_2)f_{12}(a_0,\bar a_0)=0.$ We used above the relation stated at the first step of the solution along with $ (f_{01}(a_2,\bar a_2)=f_{12}(a_0,\bar a_0)$ and $ f_{03}(a_2,\bar a_2)=-f_{23}(a_0,\bar a_0)$ which can be seen using determinants. Consequently, one can find some $ \alpha\in\mathbb{R}$ such that $ f_{01}f_{23}+f_{03}f_{12}=\alpha f_{02}f_{13}$ and in a totally similar manner, one can find some $ \beta\in\mathbb{R}$ such that $ f_{01}f_{23}-f_{02}f_{13}=\beta f_{03}f_{12}$. Combine the two to get $ (\alpha -1)f_{02}f_{13}-(\beta +1)f_{03}f_{12}=0$ and, furthermore, $ \alpha=1,\beta=-1$. This ends the proof.

I must say that this beautiful and surprising solution belongs to Calin Popescu. _____________________________________________________________________________ Moderator EDIT: Keep "Disable BBCode in this post" unchecked, in order to compile $ \color{blue}{\text{\LaTeX}}$ code.

One can easily solve the problem using the following Lemma: Let AB be a chord and P a variable point in the plane of the circle (O). If M, K and L are the feet of the perpendiculars from P onto AB and onto the tangents at A and B to the circle, then the following relation holds: PM^2 = PK*PL ( 1 ) iff P belongs to the circle (O). The proof comes directly from the similarity of the triangles PKM and PML: if P belongs to the circle, then the triangles are similar, from their similarity getting the relation (1). If the relation (1) holds, then the triangles PKM and PML are similar (s.a.s., as if KA and LB meet at T, then PKTL is cyclic and, the bisectors of the angles <ATB and <KPL are parallel; PM being perpendicular onto AB is hence the angle bisector of <KPL). But a simple angle chasing shows that these two triangles PKM and PML are similar to the triangle PAM and this compulsory brings P on the circumference ( BL is tangent to the circle (ABP), hence its center is the same with the one of (O) ). Further, apply the lemma to (ABCD), supposed cyclic, and get (ABCDE). Best regards, sunken rock

I haven't read your posts carefully, but I'm afraid Luis and sunken rock have only proved the weaker problem: Given a cyclic quadrilateral $ ABCD$ and a point $ E$ in plane, then this point $ E$ lies on the circumcircle of $ ABCDE$ if and only if $ \mathrm{d(}E,AB\mathrm{)}\cdot\mathrm{d(}E,CD\mathrm{)} = \mathrm{d(}E,AC\mathrm{)}\cdot\mathrm{d(}E,BD\mathrm{)} = \mathrm{d(}E,AD\mathrm{)}\cdot\mathrm{d(}E,BC\mathrm{)}$. The problem from the TST stated (and I don't see why this isn't clear from Ahiles' post): Let $ A$, $ B$, $ C$, $ D$, $ E$ be five given points in plane (in general position if you like). Then they are all concyclic if and only if $ \mathrm{d(}E,AB\mathrm{)}\cdot\mathrm{d(}E,CD\mathrm{)} = \mathrm{d(}E,AC\mathrm{)}\cdot\mathrm{d(}E,BD\mathrm{)} = \mathrm{d(}E,AD\mathrm{)}\cdot\mathrm{d(}E,BC\mathrm{)}$. Actually, the part you missed (the identity implies the concyclicity of $ A$, $ B$, $ C$, $ D$) is the hardest part here, in my opinion.

I proved it with (the converse of) Ptolemy's theorem. I hope it will pass as synthetic. Let $ k = \mathrm{d(}E,AB\mathrm{)}\cdot\mathrm{d(}E,CD\mathrm{)} = \mathrm{d(}E,AC\mathrm{)}\cdot\mathrm{d(}E,BD\mathrm{)} = \mathrm{d(}E,AD\mathrm{)}\cdot\mathrm{d(}E,BC\mathrm{)}$. The concyclicity of $ A$, $ B$, $ C$, $ D$ is equivalent with $ AC \cdot BD = AB \cdot CD + BC \cdot DA$. Multiplying this by $ k$, this rewrites as $ |EAC| \cdot |EBD| = |EAB| \cdot |ECD| + |EBC|\cdot |EDA|$, where $ |\mathcal{P}|$ is the unsigned area of the convex polygon $ \mathcal{P}$. Expressing these areas differently, we get the new equivalent relation: $ \sin AEC \sin BED = \sin AEB \sin CED + \sin BEC \sin DEA$ (we have deleted the term $ EA \cdot EB \cdot EC \cdot ED$). Denote $ \measuredangle AEB = x$, $ \measuredangle BEC = y$, $ \measuredangle CED = z$. In this case, the last relation can be rewritten as $ \sin (x + y) \sin (y + z) = \sin x \sin z + \sin y \sin (x + y + z)$, which can be easily verified (being true for any angles with magnitudes $ x$, $ y$, $ z$). Hence $ ABCD$ is cyclic.

Throughout a review of the proof, I've noticed that one (essential ) thing misses, that is, the fact that our conic is indeed a circle. I have reached that relation $ |a_0 - a_1|\cdot|a_2 - a_3| - |a_0 - a_2|\cdot|a_1 - a_3| + |a_0 - a_3|\cdot|a_1 - a_2| = 0$ which is equivalently to the cross ratio $ \frac {a_0 - a_1}{a_0 - a_3} : \frac {a_2 - a_1}{a_2 - a_3}$ being real and negative. But this clearly holds. It is equivalent to the vanishing of the coefficient of $ z^2$ (and $ \bar{z}^2$ as well) in (1). Hope everything's right now.

Use inversion with center E!! Let some invertion with center E be I. H1,H2,H3,H4,H5,H6 are feet of perpendicular from E to AB,CD,AC,BD,AD,BC respectively.So E,A,H1,H3,H5 E,B,H1,H4,H6 E,C,H2,H3,H6 E,D,H2,H4,H5 are concyclic. Let circle whose diameter is EA EB EC ED be C1,C2,C3,C4 respectively. Now Ci(i=1,2,3,4) move to line. We call that line li. So l1 meets l2 l3 l4 at H1' H3' H5' respectivrey. And l2 meets l3 l4 at H6' H4', l3 meets l4 at H2'.By EH1*EH2=EH3*EH4=EH5*EH6 we can get EH1'*EH2'=EH3'*EH4'=EH5'*EH6'. If E is miquel point for 4 lines, we can prove that equation by similarity. If 4 lines are given, we can decide only one E, so E is miquel point. Now we can prove that A', B', C', D' are coliner by simson's theorem(A' B' C' D' are feet of perpendicular from E to l1,l2, l3, l4 because EA EB EC ED are diameters). Therefore A,B,C,D,E are concyclic.