Two parallel lines are intersected by a transversal. Prove: Angle bisectors of the same side interior angles are perpendicular.

The answer is given below

Step-by-step explanation:

From the diagram below,Let the line AB and CD be parallel line. Let transversal line EF cut AB at Y and transversal line EF cut CD at Z.

The bisector of ∠BYZ and ∠DZY meet at O. Therefore ∠YZO = ∠DZY/2 and ∠ZYO = ∠BYZ/2

∠BYZ and ∠DZY are interior angles.

∠BYZ + ∠DZY = 180 (sum of consecutive interior angles)

∠BYZ/2 + ∠DZY/2 = 180/2

∠BYZ/2 + ∠DZY/2 = 90°

In ΔOYZ:

∠YZO + ∠ZYO + ∠YOZ = 180 (sum of angles on a straight line).

But  ∠YZO = ∠DZY/2 and ∠ZYO = ∠BYZ/2

∠DZY/2 + ∠BYZ/2 + ∠YOZ = 180

90 + ∠YOZ = 180

∠YOZ = 180 - 90

∠YOZ = 90°

Therefore Angle bisectors of the same side interior angles are perpendicular.

deez nu tz

step-by-step explanation:

you ignore the remainder, therefore the answer is y = x+2

step-by-step explanation:

you have to divide the denominator by the numerator. you can use synthetic division or long division.

[-5] .     1 .     7 .     11

x + 2 + 1/x+5