Geometry proof question (AIMO)

Question

This question is from an AIMO(Australian Intermediate Mathematics Olympiad) practise test, and I've been trying to work it out. The solution was not discussed for this question. I am looking for a complete proof.

QUESTION In the triangle ABC let $M_a$, $M_b$ and $M_c$ be the midpoints of sides $BC$, $CA$ and $AB$ respectively; and $F_a$, $F_b$ and $F_c$ be the feet of the altitudes from $A$, $B$ and $C$ respectively. Prove that:

$M_aF_b + M_bF_c + M_cF_a = F_aM_b + F_bM_c + F_cM_a$

Answer 1

The hint.

Because $$LHS=RHS=\frac{a+b+c}{2}.$$

For example $M_aF_b=\frac{a}{2}$ because $F_bMa$ is a median to the hypotenuse $BC$ of the right-angled triangle $BCF_b$.

Similarly, $F_aM_b=\frac{b}{2}.$