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Question

# In the given figure, 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 $\overline{){123}}$ QXRY is a rectangle.

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Solution

## Given : AB and CD are two parallel lines which are cut by a transversal PS at the points Q and R respectively. The bisectors of the interior angles intersect at points X and Y. To prove : Quadrilateral QXRY is a rectangle. Proof : Since AB || CD and PS is a transversal, then ∠AQR = ∠DRQ (Alternate interior angles) ⇒$\frac{1}{2}$∠AQR = $\frac{1}{2}$∠DRQ ....(1) Since QX bisects ∠AQR and RY bisects ∠DRQ, then ∠XQR = $\frac{1}{2}$∠AQR and ∠YRQ = $\frac{1}{2}$∠DRQ ∴ from (1), we get ∠XQR = ∠YRQ But ∠XQR and ∠YRQ are alternate interior angles formed by the transversal QR with QX and RY respectively. ∴ QX || RY (Alternate angles test) Similarly, we have RX || QY. Hence, in quadrilateral QXRY, we have QX || RY and RX || QY. It is known that, a quadrilateral is a parallelogram if kits opposite sides are parallel. ∴ QXRY is a parallelogram. Since sum of the interior angles on the same side of transversal is 180∘, then ∠BQR + ∠DRQ = 180∘ ⇒$\frac{1}{2}$∠BQR + $\frac{1}{2}$∠DRQ = 90∘ ....(2) Since QY bisects ∠BQR and RY bisects ∠DRQ, then ∠YQR = $\frac{1}{2}$∠BQR and ∠YRQ = $\frac{1}{2}$∠DRQ ∴ from (2), we get ∠YQR + ∠YRQ = 90∘ ....(3) In ∆QRY, we have ∠YQR + ∠YRQ + ∠QYR = 180∘ (Angle sum property of triangle) ⇒90∘ + ∠QYR = 180∘ [Using (3)] ⇒∠QYR = 180∘ − 90∘ = 90∘ Since QXRY is a parallelogram, then ∠QXR = ∠QYR (Opposite angles of ||gm are equal) ⇒∠QXR = 90∘ (∵ ∠QYR = 90∘) Since adjacent angles in a parallelogram are supplementary, then ∠QXR + ∠XRY = 180∘ ⇒90∘ + ∠XRY = 180∘ (∵ ∠QXR = 90∘) ⇒∠XRY = 180∘ − 90∘ = 90∘ Also, ∠XQY = ∠XRY = 90∘ (Opposite angles of ||gm are equal) Thus, QXRY is a parallelogram in which all the interior angles are right angles. It is known that, a rectangle is a ||gm in which each angle is a right angle. Hence, $\square$QXRY is a rectangle.

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