Question:medium

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

Show Hint

For points on a circle centered at the origin, use \[ |\vec{a}-\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2-2\vec{a}\cdot\vec{b} \] to convert chord-length conditions into dot-product relations.
Updated On: Jun 25, 2026
  • \(\vec{b}\cdot\vec{c}-\vec{a}\cdot\vec{d}=0\)
  • \(\vec{a}\cdot\vec{c}-\vec{b}\cdot\vec{d}=0\)
  • \(\vec{a}\cdot\vec{b}+\vec{c}\cdot\vec{d}=0\)
  • \(\vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{d}=0\)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Set up the circumscribed circle with center at origin.
Place the circumcircle with center at the origin. Then all four vertices satisfy $ |\vec{a}| = |\vec{b}| = |\vec{c}| = |\vec{d}| = R $, where $ R $ is the circumradius.
Step 2: Express $ AB^2 $ using vectors.
\[ AB^2 = |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 - 2\vec{a}\cdot\vec{b} + |\vec{b}|^2 = 2R^2 - 2\vec{a}\cdot\vec{b} \] This uses the identity $ |\vec{u} - \vec{v}|^2 = |\vec{u}|^2 - 2\vec{u}\cdot\vec{v} + |\vec{v}|^2 $.
Step 3: Express $ CD^2 $ using vectors.
Similarly, \[ CD^2 = |\vec{c} - \vec{d}|^2 = 2R^2 - 2\vec{c}\cdot\vec{d} \]
Step 4: Apply the given condition.
We are told $ AB^2 + CD^2 = 4R^2 $. Substituting: \[ (2R^2 - 2\vec{a}\cdot\vec{b}) + (2R^2 - 2\vec{c}\cdot\vec{d}) = 4R^2 \] \[ 4R^2 - 2(\vec{a}\cdot\vec{b} + \vec{c}\cdot\vec{d}) = 4R^2 \]
Step 5: Simplify to find the result.
\[ -2(\vec{a}\cdot\vec{b} + \vec{c}\cdot\vec{d}) = 0 \Rightarrow \vec{a}\cdot\vec{b} + \vec{c}\cdot\vec{d} = 0 \]
Step 6: State the answer.
\[ \boxed{\vec{a}\cdot\vec{b} + \vec{c}\cdot\vec{d} = 0} \]
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