Tintarn wrote: Let $ABCD$ be a convex quadrilateral with $\angle A+\angle D=90^\circ$ and $E$ the point of intersection of its diagonals. The line $\ell$ cuts the segments $AB$, $CD$, $AE$ and $ED$ in points $X,Y,Z,T$, respectively. Suppose that $AZ=CE$ and $BE=DT$. Prove that the length of the segment $XY$ is equal to the diameter of the the circumcircle of $ETZ$. Proposed by A. Kuznetsov, I. Frolov Maybe it’s “Prove that the length of the segment $XY$ is not more than the diameter of the the circumcircle of $ETZ$.” instead of “Prove that the length of the segment $XY$ is equal to the diameter of the the circumcircle of $ETZ$.” Actually 9.4 requires us to prove that the equality case can be achieved when ABCD is cyclic. I have posted 9.4 in another post.

Thank you, this is indeed the case, I corrected the statement. Sorry for messing up. By the way (for everyone), here is the link to the problem 9.4 mentioned in #2.

Let $AB$ and $CD$ intersect at $F$. Apply Menalus' Theorem twice on triangle $EZT$ with respect to lines $AB$ and $CD$ $$\frac{EB}{TB}\cdot\frac{TX}{ZX}\cdot \frac{ZA}{EA}=1,\;\;\;\; \frac{ED}{TD}\cdot\frac{TY}{ZY}\cdot \frac{ZC}{EC}=1$$Dividing out yields $XZ=TY$. Let $M$, $N$, and $O$ be the midpoints of segments $AC$, $BD$, and $XY$, respectively. Notice that $$\angle MFN+\angle MON=\angle MFN+\angle AED=90^{\circ}-\angle CAF-\angle FDB+90^{\circ}+\angle CAF+\angle FDB=180^{\circ}$$Thus $FMON$ is cyclic with diameter smaller than $FO$. Finally, $$OF\leq\frac{MN}{\sin(\angle MON)}\Rightarrow XY\leq\frac{ZT}{\sin(\angle ZET)}$$

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Tintarn wrote: Let $ABCD$ be a convex quadrilateral with $\angle A+\angle D=90^\circ$ and $E$ the point of intersection of its diagonals. The line $\ell$ cuts the segments $AB$, $CD$, $AE$ and $ED$ in points $X,Y,Z,T$, respectively. Suppose that $AZ=CE$ and $BE=DT$. Prove that the length of the segment $XY$ is not larger than the diameter of the the circumcircle of $ETZ$. Proposed by A. Kuznetsov, I. Frolov $AB \cup CD = O$. Let $\mathcal{C}$ be a hyperbola with asymptotes $AB$ and $CD$ that passes through the point $E$. Then, $\mathcal{C}$ is a rectangular hyperbola and passes through the points $Z,T$ according to the well-known property of the hyperbola and the asymptotes. By the same property, the midpoints of the $XY$ and $ZT$ segments coincide. Also, by the well-known property of a rectangular hyperbola, we understand that the point $O$ lies on the NPC of the triangle $EZT$.Therefore, the inequality that we want to prove turns into the fact that the chord of a circle is smaller than its diameter.

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Great sol @above

Let $AB$ and $CD$ meet at $P$. Note that by applying Menelaus on $EZT$ and lines $AB,CD$ we have $: \frac{AZ}{AE}.\frac{EB}{TB}.\frac{TX}{ZX} = 1 = \frac{CE}{CZ}.\frac{ZY}{TY}.\frac{TD}{TE}$ so $\frac{TX}{ZX} = \frac{ZY}{TY}$ so $XZ = TY$. Let $M,N,K$ be midpoints of $ZE,ET,TZ$. Note that diameter of $MKN$ is half the diameter of $ETZ$ and $PK$ is half the $XY$ so we need to prove $PK$ is not larger than diameter of $MNK$. Note that $\angle MPN = \angle EBP + \angle ECP - 90$ and $\angle MKN = \angle BEC = 270 - \angle EBP - \angle ECP$ so $\angle MPN + \angle MKN = 180$ so $MKNP$ is cyclic so $PK$ is at most equal to diameter of $MKN$ as wanted.