Let the circumcircle of $ABCD$ be $\odot O$ and rays $AB, DC$ intersect at ${F}$. Call the midpoints of $AC, BD, ZT$ be $P, Q, M$. Obviously $ZP=PE$ and $TQ=QE$, so ${O}$ is the circumcenter of $\triangle EZT$, so $OM \perp ZT$. By Klement Theorem $XM=YM$. Notice that $\angle AFD=\frac{\pi}{2}$, we only need to prove $FM=EO$. We will actually prove that $FEOM$ is a parallelogram. Claim 1: $FE \parallel OM$. Since ${O}$ is the circumcenter of $\triangle EZT$, we only need to prove that $EO$ and $EF$ are isogonal lines of $\angle ZET$. But actually $O, F$ are isogonal conjugates of $\triangle EAD$. Claim 2: $FE = OM$. By angle chase $\angle PFQ+\angle PMQ=\angle PFQ+\angle PEQ=\pi$, so ${F}$ is on $\odot PMQ$, the NPC of $\triangle EZT$. Combining $FE \perp ZT$, we get $FE = OM$. Done.

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 $$\angle FMO=\angle AEB+\angle FMC=90^{\circ}$$$$\angle FNO=\angle DEC+\angle BNF=90^{\circ}$$Thus $FMON$ is cyclic with diameter $FO$. Finally, $$OF=\frac{MN}{\sin(\angle MON)}\Rightarrow XY=\frac{ZT}{\sin(\angle ZET)}$$

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I believe there is a typo in the problem, it should be $BE=DT$ not $BZ=DE$. (In the grade 10 one it is written as $BE=DT$).