# 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

##### 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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