Question

IF 2 PARALLEL LINES ARE INTERSECTED BY THE TRANSVERSAL, THEN PROVE THAT BISECTORS OF THE INTERIOR ANGLES FORM A RECTANGLE.

Answer 1

Step-by-step explanation:

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.

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