How show this $\overrightarrow {AC}\cdot\overrightarrow {BD}=\frac{1}{2}[(b^2+d^2)-(a^2+c^2)]$

Question

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.

Answer 1

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

Answer 2

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 $\vec{AC}\cdot\vec{BD}={1\over2}(-c^2+b^2+d^2-a^2)$.