Question

If two parallel lines are intersected by a transversal, prove that the bisectors of the two

pairs of interior angles enclose a rectangle.​

Answer 1

Answer:

Two parallel lines AB and CD are intersected by a transversal EF in points G and H respectively. The bisectors of two pairs of interior angles intersect in L and M. To Prove: GLHM is a rectangle.... | If the sum of two adjacent angles

is 180°, then the non-common arms of the angles form a line.

Step-by-step explanation:

Two parallel lines AB and CD are intersected by a transversal EF in points G and H respectively. The bisectors of two pairs of interior angles intersect in L and M. To Prove: GLHM is a rectangle.

Proof: AB || CD and a transversal EF intersects them

.: 1/2AGH = 1/2GHD

| Alternate interior angles

ZAGH =- LACH | Halves of equals are equal

21 = 2

But these form a pair of equal alternate interior

angles

.: GM || HL .(1) Similarly, we can show that

HM || GL .(2)

In view of (1) and (2), GLHM is a parallelogram

A quadrilateral is a parallelogram if its both the

pairs of opposite sides are parallel

Now, since the sum the consecutive interior angles on the same side of a transversal is 180°

.. ZBGH +ZGHD = 180°

ZGHD =.

BGH+ | Halves of equals are equal

23 + 22 = 90° ..(3) In AGHL,

23 + 2 + ZGLH = 180° | Angle sum property of a triangle

90° + ZGLH = 180°

From (3)

ZG = 90°

- GLHM is a rectangle

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

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Answer 2

the answer for the above question is given in the attachment and the attachment consists of three pages

to see the third photo you have to slide the photos to the left to get the third one