2020 USOJMO Problems/Problem 4

Let $G$ be the intersection of $AE$ and $(ABCD)$ and $H$ be the intersection of $DF$ and $(ABCD)$ . [b][color=f00f00]Claim: $GH || FE || BC$ [/color][/b] By Pascal's on $GDCBAH$ , we see that the intersection of $GH$ and $BC$ , $E$ , and $F$ are collinear. Since $FE || BC$ , we know that $HG || BC$ as well. $\blacksquare$ [b][color=f00f00]Claim: $FB = FD$ [/color][/b] Note that since all cyclic trapezoids are isosceles, $HB = GC$ . Since $AB = BC$ and $EB \perp AC$ , we know that $EA = EC$ , from which we have that $DGCA$ is an isosceles trapezoid and $DA = GC$ . It follows that $DA = GC = HB$ , so $BHAD$ is an isosceles trapezoid, from which $FB = FD$ , as desired. $\blacksquare$

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2020 USOJMO Problems/Problem 4