Median of obtuse triangle

Question

I really cannot figure this question out. Can anyone help me please!?

Prove that the length of the median $m_a$ of obtuse triangle $△ABC$ with the obtuse $∠CAB$ is smaller than $\dfrac{1}{2}|BC|$.

Thank you very much!

Answer 1

by the theorem of cosines we have $$,m_a^2=c^2+\frac{a^2}{4}-2\frac{a}{2}c\cos(\beta)$$ with $$\cos(\beta)=\frac{a^2+c^2-b^2}{2ac}$$ we get $$m_a^2=c^2+\frac{a^2}{4}-ac\left(\frac{a^2+c^2-b^2}{2ac}\right)$$ or $$m_a^2=\frac{2(c^2+b^2)-a^2}{4}$$ and we get $$m_a^2>\frac{a^2}{4}$$ and this is equivalent to $$b^2+c^2<a^2$$ which is true for an triangle with the obtuse angle $\angle CAB$

Answer 2

Let $\Phi$ be the circle with diameter $BC$.

Since $\measuredangle BAC>90^{\circ},$ we obtain that $A$ is placed inside the circle.

Now, let $M$ be the midpoint of $BC$ with the ray $MA\cap \Phi=\{A_1\}.$

Thus, $$\frac{1}{2}BC=A_1M>AM=m_a$$ and we are done!