When two lines segments are crossed by another line segments (which is called the Transversal), the pairs of angles on opposite sides of the transversal which are outside the two lines are called Alternate Exterior Angles.

One way to easily find the alternate exterior angles is that they are the vertical angles of the alternate interior angles. Alternate exterior angles are equal to one another.

In the figure given above, angles 2 and 8 are alternate exterior angles. Angles 1 and 7 are also alternate exterior angles. Therefore, $\angle$2 = $\angle$8 and $\angle$1 = $\angle$7.

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The
couple of angles which are in the opposite sides of the transversal and
created outside the two parallel lines is determined as alternate exterior angles.

- $\angle$1 and $\angle$2 are alternate interior angles
- $\angle$a and $\angle$b are alternate interior angles

If two parallel line segments or rays are cut by a transversal, the alternate exterior angles are congruent.

**Given:**

Line p is parallel to line q and cut with the transversal l, as shown in the figure given below.

**Proof:**

S.No. | Statement | Reasons |

1 | p | | q with the transversal l | Given |

2 | $\angle$2 is congruent to $\angle$6 | Parallel Lines Postulate |

3 | $\angle$6 is congruent to $\angle$8 | Vertical Angle Theorem |

4 | $\angle$2 is congruent to $\angle$8 | Using Transitive Property |

Therefore, the alternate exterior angles are congruent.

Hence Proved.

**Step 1:** b is a supplement of 45$^o$^{}.

Therefore, b + 45$^o$^{} =180$^o$^{} => b = 180$^o$^{} - 45$^o$^{} = 135$^o$^{}

**Step 2:** b and c are vertical angles.

Therefore, c = b = 135$^o$^{}

**Step 3:** d and 45$^o$^{} are vertical angles.

Therefore, d = 45$^o$^{}

**Step 4:** d and e are alternate interior angles.

Therefore, e = d = 45$^o$^{}

**Step 5:** f and e are supplementary angles.

Therefore, f + 45$^o$^{} = 180$^o$^{} => f = 180$^o$^{} - 45$^o$^{} = 135$^o$^{}

**Step 6:** g and f are vertical angles.

Therefore, g = f = 135$^o$^{}

**Step 7:** h and e are vertical angles.

Therefore, h = e = 45$^o$

More topics in Alternate Exterior Angles | |

Alternate Exterior Angles Theorem | |

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