In the given figure, AB || CD and EF ⊥ AB. If EG is the transversal, such that ∠GED = 130°, find ∠EGF.

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Solution

We have:
$∠ C E G + ∠ G E D = 180 ° [∵ C E D is a straight line] \Rightarrow ∠ C E G + 130 ° = 180 ° \Rightarrow ∠ C E G = 50 °$
$A B ∥ C D and E G$ is the transversal.
$∠ E G F = ∠ C E G = 50 ° [Alternate Interior Angles] ∴ ∠ E G F = 50 °$

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In the given figure, AB || CD and EF ⊥ AB. If EG is the transversal, such that ∠GED = 130°, find ∠EGF.

We have: ∠CEG+∠GED=180° ∵CED is a straight line⇒∠CEG+130°=180°⇒∠CEG=50° AB∥CD and EG is the transversal. ∠EGF=∠CEG=50° Alternate Interior Angles∴∠EGF=50°

We have:
$∠ C E G + ∠ G E D = 180 ° [∵ C E D is a straight line] \Rightarrow ∠ C E G + 130 ° = 180 ° \Rightarrow ∠ C E G = 50 °$
$A B ∥ C D and E G$ is the transversal.
$∠ E G F = ∠ C E G = 50 ° [Alternate Interior Angles] ∴ ∠ E G F = 50 °$