If A, B, C and D are four points in the plane | vedclass

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(B) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively. Given that $|\vec{b}-\vec{a}|^2 + |\vec{d}-\

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(B) Let the position vectors of points $A, B, C, D$ be $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ respectively. Given that $|\vec{b}-\vec{a}|^2 + |\vec{d}-\vec{c}|^2 = |\vec{c}-\vec{b}|^2 + |\vec{a}-\vec{d}|^2$. Expanding the squares using the dot product property $|\vec{x}-\vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 - 2\vec{x} \cdot \vec{y}$: $|\vec{b}|^2 + |\vec{a}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{d}|^2 + |\vec{c}|^2 - 2\vec{c} \cdot \vec{d} = |\vec{c}|^2 + |\vec{b}|^2 - 2\vec{b} \cdot \vec{c} + |\vec{a}|^2 + |\vec{d}|^2 - 2\vec{d} \cdot \vec{a}$. Canceling common terms $|\vec{a}|^2, |\vec{b}|^2, |\vec{c}|^2, |\vec{d}|^2$ from both sides: $-2\vec{a} \cdot \vec{b} - 2\vec{c} \cdot \vec{d} = -2\vec{b} \cdot \vec{c} - 2\vec{d} \cdot \vec{a}$. Rearranging the terms: $\vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{d} - \vec{b} \cdot \vec{c} - \vec{d} \cdot \vec{a} = 0$. Factorizing the expression: $(\vec{a}-\vec{c}) \cdot (\vec{b}-\vec{d}) = 0$. This implies $\vec{AC} \cdot \vec{DB} = 0$,which means $\vec{AC} \cdot \vec{BD} = 0$.

If $A, B, C$ and $D$ are four points in the plane such that $|AB|^2+|CD|^2=|BC|^2+|DA|^2=100$,then $AC \cdot BD=$

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