Let A be the vertex of triangle ABC, wgich is divided into three equal angles x, i.e. angle A = 3x
Le AN be the altitude on BC, AM be median on BC.
Let Angle B(=y) and angle C = z.
Triangle ABN we get x+y=90 deg. .....(1)
Now triangles, ABN and ANM are congruent (ASA)
this imples, angle B = angl AMN=y
Now angle AMN = angle MAC(=x) + angle C(=z)
i.e y = x+z.......(2)
this clearly shows y>z & y> x but now we now to find relation between y & 3x.... this is where i m getting stuck.... m still trying... will come back and complete this once i solve it....

Taking triangle ANM y=90-x and in triangle ANC z=90-2x.
Also For Triangle ABN angle ABN = 90-x Therefore we can say than angle ABN = y . Thus Triangle ABM is an isocleses triangle.

Hence we can conclude AB = AM. I am not sure, but i think we can use the sine formula. y/AB= AM/y = BM/2x

I am stuck after this

no luck chayanika... m stuck too...hope someone comes to our rescue