Skip navigation.

Median length theorem

Let ABC a triangle and a=length(BC),b=length(CA),c=length(AB).
Let M be the middle of BC and D symmetric of A with respect of M.
Then ABDC is a parallelogram.

From cosine theorem we have

$$a^2=b^2+c^2-2bc \cos A$$

In triangle ABD we have


Summing up we have

$$a^2+4m_a^2=2b^2+2c^2-2bc(\cos A+\cos(\pi-A))=2b^2+2c^2$$


Average: 3.5 (8 votes)