# a simple geometric problem on pythagoras

the following figure shows the triangle $ABC$ in which $AD$ is a median and $AE$ is perpendicular to $BC$.

Prove that $2AB^2+2AC^2=4AD^2+BC^2$

Can someone help me please.I have stuck on it.

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Of ten questions you've asked so far in this site you've accepted zero... –  DonAntonio Aug 3 '13 at 12:25

By Pythagoras in the next three cases:

In triangle $\,\Delta ABE\;$

$$I\;\;\;AB^2=AE^2+BE^2$$

In triangle $\,\Delta ACE\;$

$$II\;\;\;\;AC^2=AE^2+CE^2$$

In triangle $\;\Delta ADE\;$ :

$$AD^2=AE^2+DE^2$$

Sum now $\,I+II\;$ :

$$\text{III}\;\;\;AB^2+AC^2=2AE^2+\color{red}{BE^2+CE^2}$$

But

\begin{align*}\color{red}{BE^2}&=(BD-DE)^2=BD^2-2BD\cdot DE+DE^2\\ \color{red}{CE^2}&=(CD+DE)^2=CD^2+2CD\cdot DE+DE^2\end{align*}

Thus

$$\text{III}\;\;\;AB^2+AC^2=\underbrace{2AE^2+\color{red}{2DE^2}}_{=2AD^2}+\underbrace{\color{red}{BD^2+CD^2}}_{=\frac12BC^2}+\color{red}{2DE\underbrace{(CD-BD)}_{\text{this is zero!}}}$$

And we're done...

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