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 $\boxed{123}$ QXRY is a rectangle.

Answer 1

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.