The first condition implies $ML \parallel BC$ and since $M$ is the midpoint of $AB$,$L$ is the midpoint of $AC$.Since $BL$ is the bisector of $\angle ABC$,$AM=BM=ML$,hence $BL \perp AC$,so $\bigtriangleup ABC$ is $B-$isosceles.Notice that $LB$ bisects $\angle MLH$ implies $\bigtriangleup BML \equiv \bigtriangleup BHL$,so $ML=BH$ $\implies$ $H$ is the midpoint of $BC$,which readily follows that $\bigtriangleup ABC$ is equilateral and so our third condition must also be true.

Since, $AH$ is altitude for $\Delta ABC \text{ and } \Delta HLM$, Hence, $BH||LM \implies AL=LC \implies \Delta ABC \text{ is an isosceles triangle with } AB=BC$, using the cyclic quads. we have, Let $\angle ABL=x$; $$\angle ABL=\angle AHL = \angle ACM =x \implies \angle A=\angle C=90^{\circ} -x \implies \angle BCM=90^{\circ}-2x, \text{ but } \angle BLH=x=\angle BCM \implies x=90^{\circ}-2x \implies \boxed{x=30^{\circ}}$$Hence, $\Delta ABC$ is equilateral which implies $MLCH$ is Rhombus , Hence, $CM$ bisects $HL$