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Let $DA$ be the normal on the plane of the triangle $ABC$ and $E \in (DA)$. Let's notate with $M,N,P,Q$ the proiections of the point $A$ to the lines $BD$, $CD$, $BE$, respectively $CE$. Prove that:

1) $$MN\cap BC \cap PQ \neq \emptyset;$$ 2) the quadrilateral $MNPQ$ is inscriptible.

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this is a olympiad problem and seems to be hard . Any help is very useful. I am a beginner in this kind of problem. – Iuli Sep 30 '12 at 9:37
@Iuli: I guess you mean orthogonal projection? – Christian Blatter Sep 30 '12 at 9:40
Given coordinates of A B C D E, figuring out M N P Q is not hard. Then the statements can be verified. Of course, this is an ugly, brute-force solution. – Karolis Juodelė Sep 30 '12 at 10:05
I'm probably missing something, but I don't think the first statement is true. Take for example an isosceles triangle, and then reflect the whole thing such that $B$ and $C$ get interchanged. This should leave the whole thing invariant. But this can only be the case if the lines $MN$, $BC$, $PQ$ either lie on the plane of reflection, or are perpendicular to it. Clearly none of them lie on this plane, so they must all be perpendicular to it, and hence are all parallel to each other. Since they aren't all the same, they don't intersect. – Lieven Sep 30 '12 at 11:59
up vote 2 down vote accepted

For 2), consider the circumscribed circle of the $ABC$ triangle, and expand it to a sphere in 3d with the same center and radius. If we cut it by a plane orthogonal to the $ABC$ plane and containing $AB$, we get a circle with $AB$ as its diameter, so, by Thales' thm the points $M$ and $P$ will be on that circle, hence on the sphere. Similarly for $N,Q$. Well, it's still needed that these 4 points are in the same plane..

So, for the rest, I would use vectors (or coordinate geometry), setting up the coordinate system in a preferable way, say $A$ is the origo, $ABC$ plane is the $x,y$-plane, we can set also $AB=(1,0,0)$ if it helps..

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