Question

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 ????QXRY is a rectangle.

Answer 1

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Answer 2

Answer:  Proved.

Step-by-step explanation:   Given that AB is parallel to CD and PS is a transversal.

Since the sum of interior angles on the same side of the transversal is 180°, so we have

$\angle AQR+\angle CRQ=180^\circ,~~~~~~~~~~~~~~(i)\\\\\angle BQR+\angle DRQ=180^\circ.~~~~~~~~~~~~~~(ii)$

Also, since QX, QY, RX and RY are the angle bisectors of ∠AQR, ∠BQR, ∠CRQ and ∠DRQ respectively.

So, we have

$\angle AQR=2\angle XQR,\\\\\angle BQR=2\angle YQR,\\\\\angle CRQ=2\angle XRQ,\\\\\angle DRQ=2\angle YRQ.$

Substituting these in equations (i) and (ii), we get

$2\angle XQR+2\angle XRQ=180^\circ\\\\\Rightarrow \angle XQR+\angle XRQ=90^\circ.~~~~~~~~~~~~~~(iii)$

Similarly,

$\angle YQR+\angle YRQ=90^\circ.~~~~~~~~~~~~~~~(iv)$

Since the sum of three angles of a triangle is 180°, therefore

$\angle QXR=\angle QYR=90^\circ.$

So, in quadrilateral QXRY, we have

$\angle Q+\angle X+\angle R+\angle Y\\\\=(\angle XQR+\angle YQR)+90^\circ+(\angle XRQ+\angle YRQ)+90^\circ\\\\=90^\circ+90^\circ+90^\circ+90^\circ,~\textup{using equations (iii) and (iv)}\\\\=360^\circ.$

Thus, QXRY is a quadrilateral, since the sum of four angles of a quadrilateral is 360°.