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In an acute angled triangle ABC, let H be the orthocenter, and let D, E, F be the feet of the altitudes from A, B, C to the opposite sides, respectively. Let L, M, N be the midpoints of the segments AH, EF, BC respectively. Let X, Y be feet of altitudes from L, N on to the line DF. Prove that XM is perpendicular to MY

Source of the problem
Regional Math Olympiad, 2019 Problem 5
Geometry

7/10

Suggested Book
Challenges and Thrills in Pre College Mathematics

Do you really need a hint? Try it first!

Draw a diagram.   Construction: Can you find a circle? Nine-point circle passes through feet of altitudes, the midpoint of sides and midpoints of AH, BH, CH (where H is orthocenter) of a triangle. This is a well-known theorem from geometry.  In this case, it passes through L, E, D, N, and F.

Chase three cyclic quadrilaterals to understand DH is the internal angle bisector of $\angle EDF$. Since $\angle BFE = \angle BEC = 90 ^o$ hence BFEC is cyclic hence $\angle FBE = \angle FCE$ (angle subtended by the segment FE). Since $\angle ADC = \angle AFC = 90 ^o$ hence AFDC is cyclic hence $\angle FDA = \angle FCA or \angle FCE$ (angle subtended by the segment FA). Since $\angle BDA = \angle BEA = 90 ^o$ hence BDEA is cyclic hence $\angle ADE = \angle ABE$ (angle subtended by the segment AE). Combining we have $\angle FDA = \angle EDA$ implying DA bisects $\angle FDE$ Can you show L-M-N are collinear and perpendicular to EF?

We show that $\Delta LMF$ is congruent to $\Delta LME$.  Clearly LM = LM, ME =  MF (as M is the midpoint). We will show that LE = LF. Why? Since L is the midpoint of LH and AFH is right-angled, hence L being the midpoint of hypotenuse is the circumcenter of AFH.  This implies LF = LA. Similarly, L is the circumcenter of AEH implying LE = LA.  Thus LF = LE.  Hence we showed that $\Delta LMF$ is congruent to $\Delta LME$ making $\angle LMF = \angle LME = 90^o$ Similarly NM is perpendicular to EF.  Now can you find two cyclic quadrilaterals?
Notice that FXML is cyclic ( $\angle FML = \angle FXL = 90^o$ ) Also, FMYN is cyclic ( $\angle FMN = \angle FYN = 90^o$ ) We will show $\angle FMX = \angle NMY$ (Then by adding $\angle XMN$ to both side we will be done)  Toward that end we will show triangles FXM and NYM are similar.  $\angle MFX = \angle MNY$ (subtended by MY in cyclic quadrilateral MFNY ) $\angle FXM = \angle NYM$ due to the following reason (making the triangles similar and remaining angle equal).  $\angle FXM = \angle FXL + \angle LXM = 90^o + \angle LFM = 90^o + \angle LDE (green angle)$ $\angle NYM = \angle NYF + \angle FYM = 90^o + \angle FNL = 90^o + \angle FDL (green angle)$ Hence proved.

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