Question

In an acute angled triangle ABC, let H be the orthocentre, 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.

Please answer correctly and I'll mark you as brainliest.​

Answer 1

since BE perpendicular to AC andCF perpendicular to AB.

So we have AFHE is cyclic quadrilateral also AH is diameter of this circle since L is the mid point of AH the EF is chord of circle we have LM perpendicular to EF (as M is mid point of EF).

Similarly we have BCEF is also cyclic and ′N′ is mid point of diameter of BC.Since EF is radical axis of both the circle AFHE and BCEF and L,N are centres of these circles

So,LN perpendicular to EF.

So, We have LMN are colinear.

Now LMFX,MFYN are cyclic (in both quadrilaterals sum of opposite angles=180°)

∠MLF=∠MYF=x(Let)

∠MNF=∠MYF=y(Let)

∠MY=∠LFN=90°(x+y=90°)

Since ∠LFN is angle in the semicircle of nine point circle with LN as diameter.

Hence , XM perpendicular to MY. proved