Question

Hi Expert please solve.
Prove that the sum of measure of two sides of a triangle is greater than double the median bisecting the third side.

Answer 1

Solution.

Let the triangle has sides $a, b, c$ with a median $d$ drawn to side $a$ . Let $m$ be the length of the segments of a formed by the median, so $m$ is half of $a$ . Let the angles formed between $a$ and $d$ be $\theta$ and $\theta'$ where $\theta$ includes $b$ and $\theta'$ includes $c$ . Then $\theta'$ is the supplement of $\theta$ and $\cos \theta' = -\cos \theta$ . The law of cosines for $\theta$ and $\theta'$ states

$$b ^ {2} = m ^ {2} + d ^ {2} - 2 d m \cos \theta$$ $$c ^ {2} = m ^ {2} + d ^ {2} - 2 d m \cos \theta^ {\prime} = m ^ {2} + d ^ {2} + 2 d m \cos \theta$$

Add these equations:

$$b ^ {2} + c ^ {2} = 2 m ^ {2} + 2 d ^ {2}$$

So

$$b ^ {2} + c ^ {2} - 2 m ^ {2} = 2 d ^ {2} \Rightarrow b ^ {2} + c ^ {2} \geq 2 d ^ {2}$$