If $A B C D$ is a cyclic quadrilateral with $R$ as the radius of the circumcircle and $(\mathrm{AB})^2+(C…
If $A B C D$ is a cyclic quadrilateral with $R$ as the radius of the circumcircle and $(\mathrm{AB})^2+(C D)^2=4 R^2$ then
$\bar{b} \cdot \bar{c}-\bar{a} \cdot \bar{d}=0$
$\bar{a} \cdot \bar{c}-\bar{b} \cdot \bar{d}=0$
$\bar{a} \cdot \bar{b}+\bar{c} \cdot \bar{d}=0$
$\bar{a} \cdot \bar{c}+\bar{b} \cdot \bar{d}=0$
Solution
We know that if $A B C D$ is a cyclic quadrilateral with $\mathrm{R}$ as the radius of the circumcircle and
$
\mathrm{AB}^2+\mathrm{CD}^2=4 \mathrm{R}^2
$
$\therefore \mathrm{ABCD}$ is a rhombus.
$
\begin{aligned}
& \therefore \angle \mathrm{AOB}=90^{\circ} \text { and } \angle \mathrm{COD}=90^{\circ} \\
& \therefore \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0 \text { and } \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{d}}=0
\end{aligned}
$
So. $\vec{a} \cdot \vec{b}+\vec{c} \cdot \vec{d}=0$