Question

THEOREM :If two parallel lines are intersected by a transversal, prove that the bisectors of the two
pairs of interior angles enclose a rectangle.
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Answer 1

Answer :

Proved!

Step-by-step explanation :

Given :

★ Theorem -

If two parallel lines are intersected by a transversal, prove that the bisectors of the two  pairs of interior angles enclose a rectangle.

To Prove :

The bisectors of the two  pairs of interior angles enclose a rectangle.

Solution :

★ { Refer the Attachment! }

From Figure,

The two parallel lines are X and Y and AB is the transversal.

∠BAF and BAH form -

⇒ ∠BAF + BAH = 180°

Now, Multiply the both sides by $\dfrac{1}{2}$

⇒ $\dfrac{1}{2}$ × (BAF + BAH) = 90   ..... Eq(1).

Implies,

AC is the bisector of Angle BAH

Angle BAC =  $\dfrac{1}{2}$ × (Angle BAH) -(2) × (Angle BAH)

...Eq.(2)

Now,

Addition of Equation 1 and 2 :

⇒ BAC + BAD = $\dfrac{1}{2}$ × (BAF + BAH) = 90° ...{ From Eq1 }

⇒ CAD = 90°.

Since,

⇒ CBD = 90°

Now, Quadrilateral ABCD has two opposite angles equal to 90° which is accordingly to prove that it is a rectangle.

Hence, Proved!

Answer 2

SOLUTION:-

Given:

Two parallel lines AB and CD are interested by a transversal EF in points G & H respectively.

The bisector of two pairs of interior angles intersect in L & M.

To prove:

GLHM is a rectangle.

Proof:

AB||CD and a transversal EF intersect them

Therefore,

$\angle AGH = \angle GHD \: \: \: \: (alternate \: interior \: angles) \\ \\ = > \frac{1}{2} \angle AGH = \frac{1}{2} \angle GHD \\ \\ [halves \:of \:equals\: are\: equal]\\ \\= > \angle 1 = \angle 2$

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) & (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 angels on the same side of a transversal is 180°.

Therefore,

$\angle BGH + \angle GHD = 180 \degree \\ \\ = > \frac{1}{2} \angle BGH + \frac{1}{2} \angle GHD = \frac{1}{2} (180 \degree) \: \: \: \: (halves \: of \: equals \: are \: equal) \\ \\ = > \angle 3 + \angle 2 = 90 \degree...........(3)$

In ∆GHL,

$\angle 3 + \angle 2 + \angle GHL = 180 \degree \: \: \: \: [angle \: sum \: property \: of \: a \: triangle] \\ \\ = > 90 \degree + \angle GHL = 180 \degree \: \: \: \: \: [from \: (3)] \\ \\ = > \angle GHL = 180 \degree - 90 \degree \\ \\ = > \angle GHL = 90 \degree$

So,

GLHM is a rectangle.

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

[Hence, proved]

Hope it helps ☺️