Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
513 views
in Olympiad by (6.0k points)
closed by

In an acute angled triangle ABC, let H be the orthocenter, and let D, E, F be the feet of altitudes from A, B, C to the opposite sides, respectively. Let L, M, N be midpoints of 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.

1 Answer

+1 vote
by (36.8k points)
selected by
 
Best answer

A circle having diameter BC passes through E and F. Also AEHF is a cycle quadrilateral having AH as diameter. So, EF is the common chord of two above mentioned circles which will be perpndicularily bisected by line joining their centres (which is LN). 

So L, M and N are collinear. 

Now, we claim that △LFN and △XMY are similar, because ∠MXY = ∠FLN (XLMF is cyclic quadrilateral) and ∠XYM = ∠LNF (MFNY is cyclic quadrilateral) 

Hence third angle of the two triangles will also be the same ∠XMY = ∠LFN = 90° (LN is the diameter of nine point circle passing through F) 

XM ⊥ MY

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...