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.

37/3

step-by-step explanation:

step 1: simplify both sides of the equation.

7x+8(x+

1

4

)=3(6x−9)−8

7x+(8)(x)+(8)(

1

4

)=(3)(6x)+(3)(−9)+−8(distribute)

7x+8x+2=18x+−27+−8

(7x+8x)+(2)=(18x)+(−27+−8)(combine like terms)

15x+2=18x+−35

15x+2=18x−35

tep 2: subtract 18x from both sides.

15x+2−18x=18x−35−18x

−3x+2=−35

step 3: subtract 2 from both sides.

−3x+2−2=−35−2

−3x=−37

step 4: divide both sides by -3.

−3x

−3

=

−37

−3

x=

37

3