Suppose I know that:
$AB^2=DF^2+CF^2$; $AD^2=BE^2+CE^2$; $AC^2=AE^2+CE^2$; $AC^2=AF^2+CF^2$
And also: $DF=AF-AD$ and $BE=AE-AB$
How can I prove that $AB * AE + AD * AF = AC^2$?
I start like this, but then I get stucked:
$AB*AE + AD*AF = AB*(BE+AB) + AD*(DF+AD) = AB*BE + AB^2 + AD*DF + AD^2 = AB*BE + AB^2 + AD*DF + BE^2 + CE^2 = (AE - BE)*BE + AB^2 + AD*DF + BE^2 + CE^2 = AE*BE - BE^2 + AB^2 + AD*DF + BE^2 + CE^2 = AE*BE + AB^2 + AD*DF + CE^2 = AE*(AE - AB) + AB^2 + AD*DF + CE^2 = AE^2 - AE*AB + AB^2 + AD*DF + CE^2...$
Can anyone give a hint how to continue?
Thanks!
Let $AB=x$, $BE=y$, $AD=p$, $DF=q$, now from the first $4$ equations we have \begin{eqnarray*} p^2+(x+y)^2=y^2+CE^2+(x+y)^2=y^2+AC^2 \\=y^2+(p+q)^2+CF^2=y^2+p^2+2pq+x^2 \end{eqnarray*} and so $xy=pq$.
Now \begin{eqnarray*} AB \times AE+ AD \times AF = x(x+y)+p(p+q) \end{eqnarray*} and \begin{eqnarray*} AC^2 = (x+y)^2+CE^2=(x+y)^2+p^2-y^2 \end{eqnarray*} and thus $\color{red}{AB \times AE+ AD \times AF =AC^2}$.