Theorem 15.1.3

Theorem 15.1.4 (Apollonius’s Theorem): In any triangle, the sum of the squares on any two sides is equal to twice the square on half of the third side together with twice the square on the median which bisects the third side.

Given: In $\triangle ABC$, the median $\overline{AD}$ bisects $\overline{BC}$ at point $D$. Such that $m\overline{BD} = m\overline{CD}$.

To prove: $(m\overline{AB})^2+(m\overline{AC})^2=2(m\overline{BD})^2+2(m\overline{AD})^2$

Construction: Draw $\overline{AF}\;ꓕ\;\overline{BC}$

Proof: