Question

If 2 parallel lines are intersected by a transversal, prove that the bisectors of the two pairs of interior angles encloses a rectangle.

Answer 1

AB and CD are two parallel lines intersected by a transversal L. X and Y are the points of intersection of L with AB and CD respectively. XP, XQ, YP and YQ are the angle bisectors of ∠ AXY, ∠ BXY, ∠ CYX and ∠ DYX.
AB || CD and L is transversal.
∴ ∠ AXY = ∠ DYX (Pair of alternate angles)
⇒ 1/2 ∠ AXY = 1/2 ∠ DYX
⇒ ∠ 1 = ∠ 4 (∠ 1 = 1/2 ∠ AXY and ∠ 4 = 1/2 ∠ DYX)
⇒ PX/YQ  (If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are parallel)...(1)
Also ∠ BXY = ∠ CYX  (Pair of alternate angles)
⇒ 1/2 ∠ BXY = 1/2 ∠ CYX
⇒ ∠ 2 = ∠ 3  (∠ 2 = 1/2 ∠ BXY and ∠ 3 = 1/2 ∠ CYX)
⇒ PY/XQ  (If a transversal intersects two lines in such a way that a pair of alternate interior angles are equal, then the two lines are parallel) ...(2)
From (1) and (2), we get
PXQY is a parallelogram ....(3)
∠ CYD = 180°
⇒ 1/2 ∠ CYD = 180/2 = 90°
⇒ 1/2 (∠CYX + ∠ DYX) = 90°
⇒ 1/2 ∠ CYX + 1/2 ∠ DYX = 90°
⇒ ∠3 + ∠ 4 = 90°
⇒ ∠ PYQ = 90° ...(4)
So, using (3) and (4), we conclude that PXQY is a rectangle.
Hence proved.

Answer 2

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