# 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

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.0" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" _builder_version="4.0" open="on"]Regional Math Olympiad, 2019 Problem 5[/et_pb_accordion_item][et_pb_accordion_item title="Topic" open="off" _builder_version="4.0"]Geometry

[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.29.2" open="off"]7/10

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.0" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.0"]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.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.0"]

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?

[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.0"] 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? [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.0"] 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.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.27.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]