Hello, jacs!

In $\displaystyle \Delta ABC,\:D$ is the midpoint of $\displaystyle BC$.

Prove that: .$\displaystyle AB^2 + AC^2 \:= \:2\!\cdot\!AD^2 + 2\!\cdot\!CD^2$

We have: .$\displaystyle \Delta ABC$ with median $\displaystyle AD$.

Let $\displaystyle BD = DC = x$.

Drop perpendicular $\displaystyle AE$ from vertex $\displaystyle A$ to side $\displaystyle BC$. .Let $\displaystyle h = AE$.

.Let $\displaystyle a = ED$, then: $\displaystyle BE = x - a$ Code:

A
*
*:* *
* : * *
* : * *
* : * *
* :h * *
* : * *
* : * *
B * - - - + - - - * - - - - - - - * C
x-a E a D x

In right triangle $\displaystyle AEB\!:\;\;AB^2\;=\;h^2 + (x-a)^2$ . **[1]**

In right triangle $\displaystyle AEC\!:\;\;AC^2 \;=\;h^2 + (x+a)^2$ . **[2]**

Add [1] and [2]: .$\displaystyle AB^2 + AC^2 \;=\;h^2 + (x-a)^2 + h^2 + (x+a)^2$

And we have: .$\displaystyle AB^2 + AC^2 \;=\;2(h^2+a^2) + 2x^2$ . **[3]**

. . In right triangle $\displaystyle AED\!:\;\;h^2 + a^2 \:=\:AD^2$

. . . . We also know that:.$\displaystyle x = CD$

Therefore, [3] becomes: .$\displaystyle AB^2 + AC^2 \;=\;2\!\cdot\!AD^2 + 2\!\cdot\!CD^2$