Calculate Angle MAB in a Right Triangle with 50° Base Angle

In a right triangle, the median to the hypotenuse has a special property: it equals half the length of the hypotenuse! Since $AM = \frac{1}{2}BC$ and M is the midpoint, we get $AM = BM = MC$, making triangle ABM isosceles.

In triangle ABM, since $AM = BM$, the base angles (opposite the equal sides) are equal. The base angles are $∠MAB$ and $∠MBA$, so they're both 50°.

The same principle applies! If $∠BAC = 50°$, then since triangle ACM would be isosceles with $AM = CM$, we'd have $∠MAC = ∠MCA = 50°$.

Yes! In any right triangle, the median to the hypotenuse always equals half the hypotenuse length. This creates two isosceles triangles, making angle calculations much easier.

No, this special median property only works for right triangles. In other triangles, the median to a side doesn't necessarily equal half that side's length.