How show this $\overrightarrow {AC}\cdot\overrightarrow {BD}=\frac{1}{2}[(b^2+d^2)-(a^2+c^2)]$ In Convex quadrilateral $ABCD$, such $|AB|=a,|BC|=b,|CD|=c,|DA|=d$, show that
$$\overrightarrow {AC}\cdot\overrightarrow {BD}=\dfrac{1}{2}[(b^2+d^2)-(a^2+c^2)]$$
I have one methods to solve this problem, following is my methods.
since
\begin{align*}&\overrightarrow {AB}+\overrightarrow {BC}+\overrightarrow {CD}+\overrightarrow {DA}=\overrightarrow {0}\Longrightarrow |AD|^2=(\overrightarrow {AB}+\overrightarrow {BC}+\overrightarrow {CD})^2\\
&=\overrightarrow {AB}^2+\overrightarrow {BC}^2+\overrightarrow {CD}^2+2(\overrightarrow {AB}\cdot\overrightarrow {BC}+\overrightarrow {BC}\cdot\overrightarrow {CD}+\overrightarrow {CD}\cdot\overrightarrow {AB})\\
&=a^2-b^2+c^2+2(\overrightarrow {BC}+\overrightarrow {CD})(\overrightarrow {AB}+\overrightarrow {BC})\\
&=a^2-b^2+c^2+2\overrightarrow {AC}\cdot\overrightarrow {BD}
\end{align*}
but I'm looking for other methods. It's not important to use what, only the time that it takes is important.
 A: Let $O$ be the point where the two diagonals $AC$ and $BD$ (produced if needed) meet, and set: $\vec\alpha=\vec{OA}$,   $\vec\beta=\vec{OB}$,   $\vec\gamma=\vec{OC}$,   $\vec\delta=\vec{OD}$. We have $\vec{AC}=\vec\gamma-\vec\alpha$ and  $\vec{BD}=\vec\delta-\vec\beta$, so that
$$
\tag{a}
\vec{AC}\cdot\vec{BD}=
\vec\gamma\cdot\vec\delta-\vec\beta\cdot\vec\gamma-
\vec\delta\cdot\vec\alpha+\vec\alpha\cdot\vec\beta.
$$
But by the cosine law we have 
$\vec\gamma\cdot\vec\delta={1\over2}(\gamma^2+\delta^2-c^2)$ and so on: substituting these into (a), all terms with $\alpha^2$, $\beta^2$, $\gamma^2$, $\delta^2$ cancel out and one gets 
$\vec{AC}\cdot\vec{BD}={1\over2}(-c^2+b^2+d^2-a^2)$.
A: Let $A(x_{1},y_{1}),B(x_{2},y_{2}),C(x_{3},y_{3}),D(x_{4},y_{4})$,so we have
$$\begin{cases}
(x_{1}-x_{2})^2+(y_{1}-y_{2})^2=a^2\\
(x_{2}-x_{3})^2+(y_{2}-y_{3})^2=b^2\\
(x_{3}-x_{4})^2+(y_{3}-y_{4})^2=c^2\\
(x_{4}-x_{1})^2+(y_{4}-y_{1})^2=d^2
\end{cases}$$
so we have
$$b^2+d^2-(a^2+c^2)=2(x_{1}x_{2}+x_{3}x_{4}-x_{2}x_{3}-x_{1}x_{4})+2(y_{1}y_{2}+y_{3}y_{4}-y_{4}y_{1}-y_{2}y_{3})$$
then
$$\overrightarrow{AC}\cdot\overrightarrow{BD}=x_{1}x_{2}+x_{3}x_{4}-x_{2}x_{3}-x_{1}x_{4})+y_{1}y_{2}+y_{3}y_{4}-y_{4}y_{1}-y_{2}y_{3}=\dfrac{b^2+d^2-a^2-c^2}{2}$$
