Question

please solve this question​

Answer 1

Answer

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.

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