Let $ABC$ be an arbitrary triangle. A circle passes through $B$ and $C$ and intersects the lines $AB$ and $AC$ at $D$ and $E$, respectively. The projections of the points $B$ and $E$ on $CD$ are denoted by $B'$ and $E'$, respectively. The projections of the points $D$ and $C$ on $BE$ are denoted by $D'$ and $C'$, respectively. Prove that the points $B',D',E'$ and $C'$ lie on the same circle.
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Tags: geometry, cyclic quadrilateral
motal
19.01.2011 21:21
Consider the quadrilateral $EE'CC'$. It's concyclic because $\angle{EE'C}=\angle{CC'E}=90^{\circ}$. Hence, $\angle{DCE}=\angle{E'CE}=\angle{E'C'E}=\angle{D'C'E'}$. In the same way, $B'BD'D$ is concyclic $\Longrightarrow\angle{DBE}=\angle{DBD'}=\angle{D'B'D}=\angle{D'B'E'}$. But $\angle{DCE}=\angle{DBE}$, hence $\angle{D'C'E'}=\angle{D'B'E'}$
PlatinumFalcon
27.07.2014 23:01
Since $\angle BB'C=\angle BC'C=90$, $B'C'BC$ is cyclic. Now, $E'EDD'$ is also cyclic, as $\angle EE'D=\angle ED'D=90$. Hence $\angle ED'E'=\angle B'D'E'=\angle E'DE=\angle BDE$, and $\angle B'C'E'=\angle B'C'B=\angle B'CB=\angle ECB$, and the result follows, as $EBCD$ is cyclic. $\Box$
jayme
28.07.2014 09:02
Dear Mathlinkers, this problem turns out to this result: the four projections of the vertices of a cyclic quadrilateral upon the two diagonals, are concyclics. This circle is known as the Morel's circle. Sincerely Jean-Louis
zuss77
16.06.2020 14:52
Let $X=CD\cap BE$. $B',C', D',E'$ are feet of altitudes of similar triangles $XBC$ and $XDE$. So $XB'\cdot XE'=XC'\cdot XD'$.
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parmenides51
15.08.2024 20:15
Problem is a special cass of this; ASU 047 All Russian MO 1964 9.3 Quote: Four perpendiculars are drawn from the vertices of a convex quadrangle to its diagonals. Prove that their bases make a quadrangle similar to the given one.