A(vec{a}), B(vec{b}), C(vec{c}), D(vec{d}) ar | vedclass

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(B) Given that $A(\vec{a}), B(\vec{b}), C(\vec{c}), D(\vec{d})$ are four concyclic points such that $x\vec{a}+y\vec{b}+z\vec{c}+t\vec{d}=\vec{0}$ and

Vedclass · Accepted Answer

$|xy||\vec{a}-\vec{b}|^2=|zt||\vec{c}-\vec{d}|^2$

Vedclass · Answer

(B) Given that $A(\vec{a}), B(\vec{b}), C(\vec{c}), D(\vec{d})$ are four concyclic points such that $x\vec{a}+y\vec{b}+z\vec{c}+t\vec{d}=\vec{0}$ and $x+y+z+t=0$. Since $A, B, C, D$ are concyclic and chords $AB$ and $CD$ intersect at $P$, by the power of a point theorem, $PA \cdot PB = PC \cdot PD$. Using the section formula for the intersection point $P$ of chords $AB$ and $CD$, we can express $P$ in terms of the position vectors. For chord $AB$, $P$ divides $AB$ in some ratio $k_1 : k_2$, so $\vec{p} = \frac{k_2\vec{a} + k_1\vec{b}}{k_1+k_2}$. For chord $CD$, $P$ divides $CD$ in some ratio $k_3 : k_4$, so $\vec{p} = \frac{k_4\vec{c} + k_3\vec{d}}{k_3+k_4}$. Equating these and comparing with the given linear combination $x\vec{a}+y\vec{b}+z\vec{c}+t\vec{d}=\vec{0}$, we identify the coefficients. From the geometry of intersecting chords in a circle, the product of the segments of the chords satisfies the relation $|xy||\vec{a}-\vec{b}|^2 = |zt||\vec{c}-\vec{d}|^2$. Thus, the correct option is $B$.

$A(\vec{a}), B(\vec{b}), C(\vec{c}), D(\vec{d})$ are four concyclic points, such that $x \vec{a}+y \vec{b}+z \vec{c}+t \vec{d}=\vec{0}$ and $x+y+z+t=0$, where $x, y, z, t$ are constants not all zero. If the chords $AB$ and $CD$ intersect at $P$, then:

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