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In the adjoining figure, AB || CD, t is the traversal, EG and FG are the bisectors of ∠BEE and ∠DFE respectively, then ∠EGF is equal to :
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- 90°
- 75°
- 80°
- 110°
- None of these
Correct Option: A
AB || CD and t transversal intersects them at E and F
∠BEF + ∠EFD = 180° (co-interior angles)
| ⇒ | 1 | ∠BEF | + | 1 | ∠EFD | = 180/2 |
| 2 | 2 |
⇒ ∠FEG + ∠EFG = 90°
In Δ GEF
∠EGF + ∠FEG + ∠EFG = 180°
∴ ∠EGF + 90° = 180°
∴ ∠EGF = 90°.
The above result can be restated as :
If two parallel lines are cut by a traversal, then the bisectors of the interior angles on the same side of the traversal intersect each other at right angles.
