median equals half of the side (trigonometric proof)

Question

It is well known that

If in a triangle a median has the measure half the length of the side it is drawn, then the triangle is a right triangle.

Is there any trigonometric proof of the above statement?

Answer 1

So, you have a triangle, consisting of three points, $A$ and $B$ and $D$. The midpoint of $AB$ is point $C$, from where you draw the median $DC$. As this means that the angle $ADB$ is created on a circle, it automatically means that this is half of the angle $ACD$, which is equal to $\pi$ or $180°$:

If you really want to include trigonometry, I would advise you to calculate $BD$ using the sine-rule in triangle $BCD$ and to calculate $AD$ using the sine-rule in triangle $ACD$. As the angles of $ACD$ and $BCD$ together are $\pi$ or $180°$, the relationship between the their sines will most probably reveal that $AD^2 + BD^2 = AB^2$, proving that you can use the formula of Pythagoras, proving that you have a rectangular triangle.

My question is: why do you want to prove something in such a complicated way when an "easy" solution is at hand?

Answer 2

If $\angle C$ is opposite the side $c$ that the median is drawn to, then by the Sine Law we have $$\frac{c}{\sin \angle C}=2R \implies \sin \angle C=1$$ Of course, this requires realization that the side $c$ is the diameter of the circumcircle.