Area and Length (Triangle Problem)

By considering the three triangles formed by joining $M$ to the vertices of the equilateral triangle, we see that $\tfrac12(MD + ME + MF)3\sqrt{3}$ is the area of the equilateral triangle, namely $\tfrac12(3\sqrt{3})^2 \tfrac{\sqrt{3}}{2}$ . Thus $MD + ME + MF = \boxed{\tfrac{9}{2}}$ .

Consider M exactly at centre of triangle. Draw perpendiculars from M to each side of triangle.

Since, M is in centre and ABC is equilateral...length of MD ,ME and MF will be same and they will cut each side in 2 equal parts ==> AD=AB/2

MA bisects angle A. So angle MAD becomes 30°.

Consider MA=x, so side opposite to 30° i.e. MD becomes x/2.

Apply pythagoras theorem to triangle AMD , we get AM^2=AD^2+DM^2

x^2=27/4 +(x/2)^2

Solving the above equation

x=3

MD=x/2=1.5

Since ,MD=ME=MF=1.5

MD+ME+MF= 4.5