Properties of Parallel Lines: Transitive & Symmetric Properties

Ans.Since m || l, ∠CGI = 120° (Corresponding angles)

∠FGC = 180° – 120° = 60° (Linear pair with ∠CGI)

∠BIH = ∠BFE = 40° (Corresponding angles)

∠BFE = ∠GFC = 40° (Vertically opposite angles)

Therefore, from the above properties of parallel lines,

x + 40° + 60° = 180° (Sum of angles of a triangle is 180°)

Or x = 180° – 100° = 80°

Ans. Let AB|| PQ through M point

Since PQ || RS

therefore AB || RS

Since AB || PQ and MX is the transversal

Since ∠ XMB = 45°

now ∠ BMY = ∠ MYR (AB || RS, Alternate angles)

Therefore, from the above properties of parallel lines,

∠ BMY = 40°

Adding ∠ XMB + ∠ BMY = 45° + 40°

That is, ∠ XMY = 85°

Ans. A traversal AD intersects two lines PQ and RS at points B and C respectively. Ray BE is the bisector of ∠ ABQ and ray CG is the bisector of ∠ BCS, and BE || CG

We are to prove that PQ || RS.

It is given that ray BE is the bisector of ∠ ABQ

Therefore, ∠ ABE = 1/ 2 ∠ ABQ

Similarly, ray CG is the bisector of ∠ BCS.

Therefore, ∠ BCG = 1 /2 ∠ BCS

But BE || CG and AD is the transversal

Therefore, from the above properties of parallel lines,

∠ ABE = ∠ BCG (Corresponding angles axiom)

Thus 1/ 2 ∠ ABQ = 1 /2 ∠ BCS

That is, ∠ ABQ = ∠ BCS

But, they are the corresponding angles formed by transversal AD with PQ and RS; and are equal. Therefore, PQ || RS (Converse of corresponding angles axiom)

Ans. Given AB || CD and BE is the transversal

So x= y (corresponding angles) (1)

Also, CD || EF and BE is the transversal

Therefore, from the above property of parallel lines,

y + 55 degrees = 180 degrees

y = 180 -55

y = 125 degrees

from (1) x = 125 degrees

Ans. We know that a linear pair is equal to 180°.

So, x+50° = 180°

∴ x = 130°

We also know that vertically opposite angles are equal.

So, y = 130°

In two parallel lines, the alternate interior angles are equal. In this

x = y = 130°

Thus from the above properties of parallel lines, it proves that alternate interior angles are equal and so, AB ||CD.

Ans. It is known that AB ||CD and CD||EF

As the angles on the same side of a transversal line sums up to 180°,

x + y = 180° —–(i)

Also,

O = z (Since they are corresponding angles)

y +O = 180° (Since they are a linear pair)

So, y+z = 180°

let y = 3w and hence, z = 7w (As y : z = 3 : 7)

From the above property of parallel lines, 

∴ 3w+7w = 180°

Or, 10 w = 180°

So, w = 18°

Now, y = 3×18° = 54°

and, z = 7×18° = 126°

Now, angle x can be calculated from equation (i)

x+y = 180°

Or, x+54° = 180°

∴ x = 126°

Ans. Since AB ||CD, GE is a transversal.

It is given that GED = 126°

So, GED = AGE = 126° (As they are alternate interior angles)

Also,

GED = GEF +FED

As EF⊥ CD, FED = 90°

From the above properties of parallel lines, 

∴ GED = GEF+90°

Or, GEF = 126° – 90° = 36°

Again, FGE +GED = 180° (Transversal)

Putting the value of GED = 126° we get,

FGE = 54°

So,

AGE = 126°

GEF = 36° and

FGE = 54°

Ans. From the diagram,

APQ = PQR (Alternate interior angles)

Now, putting the value of APQ = 50° and PQR = x we get,

x = 50°

Also,

APR = PRD (Alternate interior angles)

Or, APR = 127° (As it is given that PRD = 127°)

From the above property of parallel lines, we know that

APR = APQ+QPR

Now, putting values of QPR = y and APR = 127° we get,

127° = 50°+ y

Or, y = 77°

Thus, the values of x and y are calculated as:

x = 50° and y = 77°

Ans. First, draw two lines BE and CF such that BE ⊥ PQ and CF ⊥ RS.

Now, since PQ || RS,

So, from the above property of parallel lines,  BE || CF

We know that,

The angle of incidence = Angle of reflection (By the law of reflection)

So,

1 = 2 and

3 = 4

We also know that alternate interior angles are equal. Here, BE ⊥ CF and the transversal line BC cuts them at B and C

So, 2 = 3 (As they are alternate interior angles)

Therefore, from the above property of parallel lines,

1 +2 = 3 +4

Or, ABC = DCB

So, AB ||CD (alternate interior angles are equal)