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.

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
$$\overrightarrow{AC}\cdot\overrightarrow{BD}=x_{1}x_{2}+x_{3}x_{4}-x_{2}x_{3}-x_{1}x_{4})+y_{1}y_{2}+y_{3}y_{4}-y_{4}y_{1}-y_{2}y_{3}=\dfrac{b^2+d^2-a^2-c^2}{2}$$
