In figure 2.28, line PS is a transversal of parallel line AB and line CD. If Ray QX, ray QY, ray RX, ray RY are angle bisectors, then prove that is a rectangle.

Given: PS is transversal of parallel line AB and line CD.

To find: QXRY is rectangle.

∠ AQR + ∠ CRQ = 180°

(divide by 2)

∠ XQR + ∠ XRQ = 90°

[ QX and RX are bisector)

In Δ XQR

∠ XQR + ∠ XRQ + ∠ QXR = 180°

90° + ∠ QXR = 180° (∠ XQR + ∠ XRQ = 180° proved above)

∠ QXR = 180° -90°

∠ QXR = 90°

Similarly, ∠ QYR = 90°

∠ AQR + ∠ BQR = 180 (straight line)

(divide by 2)

∠ XQR + ∠ YQR = 90° (QX and QY are bisector ∠)

∠ XQY = 90°

Similarly, ∠ XRY = 90°

If any quadrilateral has all the angle 90° it is a rectangle, so that QXRY is rectangle.