Olympiad1

Olympiad problem

Consider two interesecting lines BA, BC and a segment DE, joining two points of these lines, respectively. Draw a third line FG, cutting BA at G and BC at F, respectively. Calculate the angle between the lines FG and RT, where R, T are the middles of DE and FG, respectively. Show that this is a right angle, exactly when BN passes through the circumcenter of the triangle BFG. BN is the radical axis of the circumcircles of the two triangles: BDF and BEG.

Proof by Antreas Varverakis.
-Consider the circumcenters K, J of the triangles BDF and BEG, respectively. Let DM, EL be diameters of these circles, respectively. The following remarks conclude to a proof.
- Orthogonal triangles EGL, DFM are similar, the angles at M, L being equal to that at B.
- As E, D vary on lines BC, BA respectively, points M, L move on lines BM, BL, orthogonal to BA, BC respectively.
- Triangles EFN, GND are similar.
- Triangles FND, ENG are similar.
- Triangle RST is similar to the above FND, ENG.
- Angles FND, ENG and LBM are equal (pi-beta).
- The angles SRT = NEG = NBG and STR = NDF = NBE.
- Angle RTG = UTG + RTU = BFG+NBG.
- RT orthogonal to FG <= > RTU = NBG = pi/2 - f <= > BN passes through the circumcenter of BFG.
- E, D project on ML to the intersection-points of ML with the circles.
When the middle R of ED is on the medial line of the triangle's basis FG, then the line ED is tangent to an interesting parabola, studied in MedialParabola.html .

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