In the acute-angled triangle $ABC$ on the sides $AC$ and $BC$, points $D$ and $E$ are chosen such that points $A, B, E$, and $D$ lie on one circle. The circumcircle of triangle $DEC$ intersects side $AB$ at points $X$ and $Y$. Prove that the midpoint of segment $XY$ is the foot of the altitude of the triangle, drawn from point $C$.
Problem
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Tags: geometry, altitude, Kharkiv, midpoint, Concyclic
wassupbros
11.03.2020 18:57
Not every time does the circumcircle of triangle DEC intersects side AB. You need some conditions
Mario_Leopoldo
11.03.2020 20:04
It just means that $CO \perp AB$ (not difficult!). $O$: circumcenter of triangle $DEC$.
DarthVader2618
14.03.2020 14:15
Proof:
Let $M$ be the midpoint of $XY$ and the $(DEC)$ be $W$
Power of A w.r.t $W$={$AM$+$MX$}×{$AM$-$MX$}=$AC$($AC$-$DC$).
Therefore $AM^2$-$MX^2$=$AC^2$-$AC$×$DC$ eqn 1
Similarly
Power of B w.r.t $W$= $BM^2$-$MY^2$=$BC^2$-$BC$×$EC$ eqn 2
Let $(ADEB)$ be $R$
Power of C w.r.t $R$=$CD$×$CA$=$CE$×$CB$. eqn 3
Eqn1-Eqn2= $AM^2$-$BM^2$=$AC^2$-$BC^2$. Eqn4
Let $CM'$ be perpendicular to $AB$
Applying Pythagoras in $AM'C$ &$BM'C$
$AC^2$-$BC^2$=$AM'^2$-$BM'^2$
Eqn5
Comparing equations 4 and 5, we obtain that $M$ & $M'$ must coincide hence $CM$ is perpendicular to $AB$