Chapter 6: Lines and Angles

According to the question,

In figure,

OD ⊥ OE,

OD and OE are the bisectors of ∠AOC and ∠BOC.

To prove: Points A, O and B are collinear

i.e., AOB is a straight line.

Proof:

Since OD and OE bisect angles ∠AOC and ∠BOC, respectively.

∠AOC = 2∠DOC …(eq.1)

And ∠COB = 2∠COE …(eq.2)

Adding (eq.1) and (eq.2), we get

∠AOC = ∠COB = 2∠DOC + 2∠COE

∠AOC +∠COB = 2(∠DOC +∠COE)

∠AOC + ∠COB = 2∠DOE

Since, OD⊥OE

We get,

∠AOC +∠COB = 2×90^o

∠AOC +∠COB =180^o

∠AOB =180^o

So, ∠AOC + ∠COB form linear pair.

Therefore, AOB is a straight line.

Hence, points A, O and B are collinear.

According to the question,

We have from the figure ∠1 = 60° and ∠6 = 120°

Since, ∠1 = 60° and ∠6 = 120°

Here, ∠1 = ∠3 [since they are vertically opposite angles]

∠3 = ∠1 = 60°

Now, ∠3 + ∠6 = 60° + 120°

⇒ ∠3 + ∠6 = 180°

We know that,

If the sum of two interior angles on the same side of l is 180°, then the lines are parallel.

Therefore, m || n

l || m and t is the transversal

∠MAB = ∠SBA [alternate angles]

⇒ ½ ∠MAB = ½ ∠SBA

⇒ ∠PAB = ∠QBA

⇒ ∠2 = ∠3

But, ∠2 and ∠3 are alternate angles.

Hence, AP||BQ.

AP is the bisector of ∠MAB

BQ is the bisector of ∠SBA.

Given: AP||BQ.

As AP||BQ,

We have,

So ∠2 = ∠3 [Alternate angles]

2∠2 = 2∠3

⇒ ∠2 + ∠2 = ∠3 +∠3

From figure, we have ∠1= ∠2and ∠3 = ∠4

⇒ ∠1+ ∠2 = ∠3 +∠4

⇒ ∠MAB = ∠SBA

But, we know that these are alternate angles.

Hence, the lines l and m are parallel, i.e., l ||m.

Construction:

Extend DE to intersect BC at point, P.

Given, EF||BC and DP are the transversal,

∠DEF = ∠DPC …(eq.1) [Corresponding angles]

Also given, AB||DP and BC is the transversal,

∠DPC = ∠ABC …(eq.2) [Corresponding angles]

From (eq.1) and (eq.2), we get

∠ABC = ∠DEF

Hence, Proved.