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

Question

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.

Answer 1

Area can found by: $\frac{1}{2}ab\sin\theta$ ; where $\theta$ is the angle between $a$ and $b$

$$\text{Area of $\Delta$ABC}=\text{$2\times$ Area of $\Delta$MBC}$$ $$\require{cancel}\frac{1}{2}\cdot\bcancel{\text{AD}}\cdot\text{BC}=2\times\frac{1}{2}\cdot\bcancel{\text{BM}}\cdot\text{BC}\cdot(\sin\angle\text{MBC})\tag{$\because$ AD=BM}$$ $$\fbox{$\therefore\angle\text{MBC}=30^o$}$$

Answer 2

Let $MK$ be an altitude of $\Delta MBC$.

Thus, $$MK=\frac{1}{2}BM,$$ which says $$\measuredangle MBC=30^{\circ}.$$