Geometry - Vectors. Two perpendicular lines in a circle.

Question

There are two perpendicular lines in a circle. Let the middle of the circle be O, the lines being AB and CD. The two intersect at M point. Prove that OA+OB+OC+OD=2OM. (all of them being vectors).

Answer 1

Let the unit vectors along the directions $AB$ and $BC$ be $\hat{i}$ and $\hat{j}$ respectively.

Let the distances of lines $AB$ and $CD$ from the center $O$ be $p$ and $q$ respectively.

The coordinates of $M$ are thus $(\pm q, \pm p) \implies OM = \pm q\hat{i}\pm p\hat{j}$

$OA+OB$ is directed along the perpendicular bisector of $AB$, and has a magnitude twice as large as the distance between $O$ and $AB$. Hence, $OA+OB=\pm2p\hat{j}$

$OC+OD$ is directed along the perpendicular bisector of $CD$, and has a magnitude twice as large as the distance between $O$ and $CD$. Hence, $OC+OD=\pm2q\hat{i}$

$\implies OA + OB + OC + OD = \pm2q\hat{i}\pm2p\hat{j} = 2OM$