Here we will solve the problem on inequalities in triangle.
Let XYZ be a triangle in which XM bisects ∠YXZ.
Prove that XY is greater than YM.
As XM bisects ∠YXZ, we have ∠YXZ = ∠MXZ ………… (i)
Also, in ∆XMZ, ∠XMY > ∠MXZ, as an exterior angle of a
triangle is always greater then either of the interior opposite angles.
Therefore, ∠XMY > ∠YXM, [From (i)].
Therefore, XY > YM, as the greater angle has the greater
side opposite to it.