In figure 2.28, line PS is a transversalof parallel line AB and line C

1.	In figure 2.28, line PS is a transversalof parallel line AB and line CD. If RayQX, ray QY, ray RX, ray RY are anglebisectors, then prove that QXRY is arectangle.
Answer» Given : Two parallel LINES AB and CD are intersected by a Transversal PR in points Q and R respectively. The bisectors of two PAIRS of interior angles intersect in Y and X. To prove: QXRY is a rectANGLE. Proof : $AB \parallel CD$ and a Transversal QR intersects them < AQR = <QRD ( Alternate interior angles ) $\implies \frac{1}{2} \angle {AQR } = \frac{1}{2} \angle {QRD}$ ( Halves of equals are equal ) => <1 = <2 But these form a PAIR of equal alternate interior angles. $\therefore QX \parallel RY \: ---(1)$ Similarly , we can show that $RX \parallel QY \: ---(2)$ /* From (1) and (2) , / QYRX is a parallelogram.* Now, The sum of consecutive interior angles on the same side of a Transversal is 180° . $\angle {BQR} + \angle {QRD} = 180\degree$ $\implies \frac{1}{2} \angle {BQR} +\frac{1}{2} \angle {QRD} =\frac{1}{2} \times 180\degree$ $\implies \angle 3 + \angle 2 = 90\degree \: --(3)$ In ∆QRY , $\angle 3 + \angle 2 + \angle {QRY} = 180\degree$ /* Angle sum Property / $\implies 90\degree + \angle {QRY} = 180\degree \: [ From \: (3) ]$ $\angle {QRY} = 90\degree$ $\implies QYRX \: is \:a \: <klux>RECTANGLE</klux> .$ / A parallelogram with one of its angles of measure 90° is a rectangle. */ •••♪

In figure 2.28, line PS is a transversalof parallel line AB and line CD. If RayQX, ray QY, ray RX, ray RY are anglebisectors, then prove that QXRY is arectangle.

Answer»

Given :

Two parallel LINES AB and CD are intersected by a Transversal PR in points Q and R respectively.

The bisectors of two PAIRS of interior angles intersect in Y and X.

To prove:

QXRY is a rectANGLE.

Proof :

$AB \parallel CD$

and a Transversal QR intersects them

< AQR = <QRD

( Alternate interior angles )

$\implies \frac{1}{2} \angle {AQR } = \frac{1}{2} \angle {QRD}$

( Halves of equals are equal )

=> <1 = <2

But these form a PAIR of equal alternate interior angles.

$\therefore QX \parallel RY \: ---(1)$

Similarly , we can show that

$RX \parallel QY \: ---(2)$

/* From (1) and (2) , */

QYRX is a parallelogram.

Now, The sum of consecutive interior angles on the same side of a Transversal is 180° .

$\angle {BQR} + \angle {QRD} = 180\degree$

$\implies \frac{1}{2} \angle {BQR} +\frac{1}{2} \angle {QRD} =\frac{1}{2} \times 180\degree$

$\implies \angle 3 + \angle 2 = 90\degree \: --(3)$

In ∆QRY ,

$\angle 3 + \angle 2 + \angle {QRY} = 180\degree$

/* Angle sum Property */

$\implies 90\degree + \angle {QRY} = 180\degree \: [ From \: (3) ]$

$\angle {QRY} = 90\degree$

$\implies QYRX \: is \:a \: <klux>RECTANGLE</klux> .$

/* A parallelogram with one of its angles of measure 90° is a rectangle. */

•••♪

In figure 2.28, line PS is a transversalof parallel line AB and line CD. If RayQX, ray QY, ray RX, ray RY are anglebisectors, then prove that QXRY is arectangle.

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In figure 2.28, line PS is a transversalof parallel line AB and line CD. If RayQX, ray QY, ray RX, ray RY are anglebisectors, then prove that QXRY is arectangle.​

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In figure 2.28, line PS is a transversalof parallel line AB and line CD. If RayQX, ray QY, ray RX, ray RY are anglebisectors, then prove that QXRY is arectangle.