In ∆ ABC, the altitude AD and the median BM are equal in length and they intersect inside ∆ ABC. If ABC is not an equilateral triangle, ∠MBC is I have got no clue on how to proceed with the question as nothing is given except Median of one side =altitude of another moreover they also removed the possibility of equilateral triangle.


Area can found by: $\frac{1}{2}ab\sin\theta$ ; where $\theta$ is the angle between $a$ and $b$
$$\text{Area of $\Delta$ABC}=\text{$2\times$ Area of $\Delta$MBC}$$
$$\require{cancel}\frac{1}{2}\cdot\bcancel{\text{AD}}\cdot\text{BC}=2\times\frac{1}{2}\cdot\bcancel{\text{BM}}\cdot\text{BC}\cdot(\sin\angle\text{MBC})\tag{$\because$ AD=BM}$$
$$\fbox{$\therefore\angle\text{MBC}=30^o$}$$
