# Understand the problem

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

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# Start with hints

**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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