Answer:
The bisectors form a rectangle FPEQ as shown in the diagram in the attachment.
Explanation:
Please see the attachment below for an illustrative diagram.
In the diagram, lines AB and CD are the two parallel lines and XY is the transversal.
The bisectors are PE, QE, PF, and QF.
∠BPY , ∠DQX, ∠APY, and ∠CQX are the interior angles
Sum of interior angles formed by a transversal cutting through two parallel lines is supplementary.
That is,
∠BPY + ∠DQX = 180°
and
∠APY + ∠CQX = 180°
Considering ∠BPY and ∠DQX,
of ∠BPY +
of ∠BPY +
In the diagram
of ∠BPY = a
and
of ∠DQX = b
∴ a + b = 90°
NOTE: PEQ is a triangle with angles a, b and c
∴ a + b + c = 180°
Recall, a + b = 90°
Then, 90° + c = 180°
c = 180° - 90°
c = 90°
c = ∠PEQ
∴ ∠PEQ = 90°
If c is 90°, then using the same approach, d is also 90°
d = ∠PFQ
∴ ∠PFQ = 90°
Now, consider straight line CQD,
∠CQD = 180° (Sum of angles on a straight line)
That is,
∠CQX + ∠DQX = 180°
Then,
of ∠CQX +
of ∠CQX +
From the diagram,
of ∠CQX = f
of ∠DQX = b
∴ f + b = 90°
f + b = ∠FQE
∠FQE = 90°
Using the same approach
∠FPE = 90°
Now,
∠APY = ∠DQX (Alternate angles are equal)
of ∠APY =
of ∠DQX
∴ e = b
Since e = b, then lines FP and QE are parallel
Using the same approach, lines FQ and PE are parallel
Since FP and QE are opposite sides and parallel
and FQ and PE and opposite and parallel
Then, we have a parallelogram ( a quadrilateral with both pairs of opposite sides parallel).
The angles formed by the bisectors are
∠FPE, ∠PEQ, ∠FQE, and ∠PFQ which are the angles formed by the parallelogram.
Each of angles ∠FPE, ∠PEQ, ∠FQE, and ∠PFQ are equal to 90° as proven above.
A parallelogram which has each of its angles equal 90° is a rectangle.
The rectangle in the diagram is rectangle FPEQ