If ABCD is a cyclic quadrilateral with R as t | vedclass

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(C) Let the circumcircle be centered at the origin $O$ with radius $R$. The vertices $A, B, C, D$ are represented by vectors $\vec{a}, \vec{b}, \vec{c

Vedclass · Accepted Answer

$\vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{d} = 0$

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(C) Let the circumcircle be centered at the origin $O$ with radius $R$. The vertices $A, B, C, D$ are represented by vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ such that $|\vec{a}| = |\vec{b}| = |\vec{c}| = |\vec{d}| = R$. Given $(AB)^2 + (CD)^2 = 4R^2$. Since $(AB)^2 = |\vec{b} - \vec{a}|^2 = |\vec{b}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} = 2R^2 - 2\vec{a} \cdot \vec{b}$ and $(CD)^2 = |\vec{d} - \vec{c}|^2 = 2R^2 - 2\vec{c} \cdot \vec{d}$. Substituting these into the given equation: $(2R^2 - 2\vec{a} \cdot \vec{b}) + (2R^2 - 2\vec{c} \cdot \vec{d}) = 4R^2$. $4R^2 - 2(\vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{d}) = 4R^2$. $-2(\vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{d}) = 0$. Therefore,$\vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{d} = 0$.

If $ABCD$ is a cyclic quadrilateral with $R$ as the radius of the circumcircle and $(AB)^2+(CD)^2=4R^2$,then:

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