Let $ABCD$ be a square and let $X$ be any point on side $BC$ between $B$ and $C$. Let $Y$ be the point on line $CD$ such that $BX = YD$ and $D$ is between $C$ and $Y$ . Prove that the midpoint of $XY$ lies on diagonal $BD$.
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Tags: midpoint, square, geometry
L567
11.01.2021 06:45
I think the question is supposed to be prove that the midpoint of $XY$ lies on $AC$. $BX = YD \implies CX = CY$ and so $\angle XYC = \angle YXC = 45^\circ$, since $\angle ACD = \angle ACB = 45^\circ$ as well, we get that $AC \perp XY$, since $CXY$ is an isosceles triangle, the midpoint coincides with the foot of the altitude and so the midpoint lies on $AC$
lifeismathematics
05.11.2022 17:58
mark $XY\cap BD=P, YD=x ,DC=a$ then we get $XC=a-x$ and $BC=x$
from lotus lemma:-https://artofproblemsolving.com/community/q1h509770p25873315
we get $\frac{YP}{YX}=\frac{x}{a}(1+\frac{a-x}{x})$,
$\frac{YP}{YX}=\frac{x}{a}\cdot \frac{a}{x}=1$
so $YP=YX$ and hence midpoint of $XY \implies "P"$ lies on $BD$. $\blacksquare$