We have ABIIED and BC||EF
(i) BC is transversal
∠DGC = ∠ABC [corresponding angles]
But ∠ABC = 70°
∴ ∠DGC = 70°
(ii) ED is a transversal to BC|lEF
∴ ∠DEF = ∠DGC [corresponding]
∠DGC = 70°
∠DEF = 70°

LEAB = 60° and LACD = 105°
Determine
(i) LCAF and
(ii) LBAC

(i) Since EF II GH and AC is a transversal
⇒ LCAF + LACH = 180
⇒ LCAF + 105° = 180°
=75°
(ii) ∴ EF II GH and AC is transversal.
∴ LEAC = LACH
⇒ LBAC = 105°
⇒ LBAC + LBAB = 105°
⇒ LBAC + 60° = 105°
⇒ LBAC = 105°- 60°
= 45°
∴ LCAF = 75° and LBAC = 45°

No, two lines cannot intersect in more than one point.

Let the angles be 3x and 2x
Since they are linear pair of angles, their sum is 180°.
Therefore, 

x = 36°
The angles are 3x = 3 x 36 = 108°
2x = 2 x 36 = 72°

We know that \(\angle\)POR = \(\angle\)POQ + \(\angle\)QOR
62° = 23° + \(\angle\)QOR
Subtracting 23° on both sides
62° – 23° = 23° + \(\angle\)QOR – 23°
\(\angle\)QOR = 39°

∠COB = 50°
∠AOD = 50° (vertically opposite angles)
Now ∠AOC and ∠COB form a linear pair
Thus ∠AOC + ∠COB = 180°
⇒ ∠AOC + 50° = 180°.
∠AOC = 180° - 50° = 130°
Also ∠AOC and ∠BOD are vertically opposite angles
∠BOD = ∠AOC = 130°
Thus the three angles are
∠AOD = 50°
∠AOC = 130°
∠BOD = 130°

∠BAD = ∠BAE + ∠EAD
= 40° + 30° = 70°.
and LCDA = 70°
∠BAD = ∠CDA
But they form a pair of alternate angles
⇒ AB || CD
Also ∠BAE + ∠AEF = 40° + 140° = 180° -------(1)
But they form a pair of interior opposite angles.
⇒ AB II EF
From (1) and (2), we get ---(2)
AB || CD || EF
⇒ CD II EF

Step 1: Draw the given angle ∠ABC with the measure 80° using protractor.

Step 2: With B as center and convenient radius, draw an arc to cut BA and BC. Mark the points of intersection as E on BA and F on BC.

Step 3: With the same radius and E as center, draw an arc in the interior of ∠ABC and another arc of same measure with center at F to cut the previous arc.

Step 4: Mark the point of intersection as G. Draw a ray BX through G. BG is the required bisector of the given angle ∠ABC

Given l is parallel to m and n is transversal to l and m.
We get, y = 2x [Vertically opposite angles are equal]
y + 4x = 180° [sum of interior angles that lie on the same side of the transversal]
2x + 4x = 180° [since y = 2x]
6x = 180°
Dividing by 6 on both sides
\(\frac{x}{6}=\frac{180^o}{6}\)gives, x = 30°
Now, y = 2(30°) = 60°.

(i) The angle that corresponds to \(\angle\)1 is \(\angle\)5
(ii) The angle that is alternate interior to \(\angle\)3 is \(\angle\)5
(iii) The angle that is alternate exterior to \(\angle\)8 is \(\angle\)2
(iv) The angle that corresponds to \(\angle\)8 is \(\angle\)4
(v) The angle that is alternate exterior to \(\angle\)7 is \(\angle\)1
(vi) The angle that is alternate interior to \(\angle\)6 is \(\angle\)4