From above given triangle , It is clear that the altitudes AL, BM and CN of ΔABC intersect at H. Then H is the orthocentre of ΔABC.
In ΔABC, HL ⊥ BC and BN ⊥ CH.
Thus, the two altitudes HL and BN of ΔHBC, intersects at A.
Hence the orthocentre of ΔHBC is M .