nice [midpoints in Simson line configuration: PQ perp. DE]

[hossein11652]

Let M be a point on the circumcircle of ABC. and D and E be the feet of the perpendeculars from M to BC and AC. Let P and Q be the midpoints of AD and BE. prove that PQ is perpendecular to DE.

[Omid Hatami]

Well, The official solution is that suppose N is midpoint of AB.
Two triangles MDE and PQN are similiar and their sides are pependicular.
But I solved it in a beutiful way.
Suppose M' is reflection of M in line BC.
Because of well-known theorem,HM' is pararell to DE(H orthocenter)
We can easily prove that HM' is radial axis of two circles with diameters AD and BE.
And problem has been solved

[darij grinberg]

What should I say? It's a problem on triangle geometry, hence I must answer ;D , but the most "elementary" solution I have needs three lemmas, so let me just hope you know these lemmas and won't have problems with the lots of preknowledge used.

My first lemma is rather harmless:

Lemma 1, the Simson line theorem. If M is a point on the circumcircle of triangle ABC, then the feet D, E, F of the perpendiculars from M to the sides BC, CA, AB lie on one line, the so-called SIMSON LINE s of the point M with respect to triangle ABC.

Hence, the line DE is nothing but our Simson line s. We must show that this Simson line s is perpendicular to the line PQ, where P and Q are the midpoints of the segments AD and BE.

Note that the line s is not only the Simson line of the point M with respect to triangle ABC, but it is also the Simson line of M with respect to the triangles AEF, BFD, CDE. This is because the point M lies on the circumcircles of these triangles (in fact, from < AEM = 90� and < AFM = 90�, we see that the points A, E, F, M are concyclic, i. e. the point M lies on the circumcircle of triangle AEF, and similarly it lies on the circumcircles of triangles BFD and CDE), and the feet of the perpendiculars from the point M to the sides of these triangles trivially lie on s (to be honest, each of these triangles just has a sideline coinciding with s).

Lemma 1 helped us to identify the line DE as the Simson line s. Another lemma will give us some clues about PQ:

Lemma 2, the Steiner and Gauss lines theorem. Four lines form four triangles. The orthocenters of these four triangles lie on one line, the so-called STEINER LINE of the complete quadrilateral formed by our four lines. Also, the midpoints of the diagonals of this complete quadrilateral lie on another line, the so-called GAUSS LINE of the complete quadrilateral. The Gauss line and the Steiner line are perpendicular to each other.

In our case, if we consider the complete quadrilateral formed by the lines BC, CA, AB and the Simson line s, then the diagonals of this quadrilateral are the segments AD, BE, CF. The midpoints of the first two diagonals are P and Q. Hence, the line PQ is the Gauss line of our complete quadrilateral.

Therefore, we can rewrite the assertion we must prove as follows: The Simson line s is perpendicular to the Gauss line of the complete quadrilateral formed by the lines BC, CA, AB and s. Now, from Lemma 2, this Gauss line is perpendicular to the Steiner line of this quadrilateral. Hence, it remains to prove that the Simson line s is parallel to the Steiner line of the complete quadrilateral formed by the lines BC, CA, AB and s.

Well, by definition, this Steiner line passes through the orthocenters of triangles ABC, AEF, BFD, CDE. Now a last (well-known?) lemma will help:

Lemma 3. The Simson line s of the point M with respect to triangle ABC bisects the segment MH, where H is the orthocenter of triangle ABC.

Thus, the image of the Simson line s in the homothety with center M and factor 2 is a line s' passing through the orthocenter H of triangle ABC. Now, since s is also the Simson line of the point M with respect to the triangles AEF, BFD, CDE, this image line s' must also pass through the orthocenters of the triangles AEF, BFD, CDE. Altogether, the line s' contains the orthocenters of the four triangles ABC, AEF, BFD, CDE. Consequently, the line s' is the Steiner line of the complete quadrilateral formed by the lines BC, CA, AB and s. Now, since the line s' is the image of the line s in the homothety with center M and factor 2, the line s' is parallel to s. So the Steiner line of our complete quadrilateral is indeed parallel to the Simson line s. Proof complete.

Darij

[Omid Hatami]

The official solution is shorter and uses nothing except similiarity of 2 triangles.

[darij grinberg]

Indeed, but anyone with the triangle geometry know-how immediately sees my solution, while the idea of the proposed one is not so immediate...

Darij

[sam-n]

I use complex number to solve it on unit circle ,really IQ free solution

[Pascual2005]

well i knew steiner line, but Gauss Line? whats that, could you explain what it is a little bit.

Thanks

[hossein11652]

Well done Omid
It was very nice

[Omid Hatami]

Thank you very much Amir.
but to Pascual2004 Suppose four lines
( intersection of and )
Then midpoint of these segments are collinear

[darij grinberg]

[treegoner]

I don't think this one is complex. Using vector and Simson line should help.
Call K the intersection of DE and AB.
We have
Thus
which is equivalant to (obvious).
Anyway, Darij 's performance reveals that he knows a lot in geometry.