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Alternate and Corresponding AnglesDate: 10/21/96 at 19:20:0 From: Wesley Battersby Subject: Geometry Please explain corresponding and alternate angles. Thanks, Wesley B.
Date: 02/11/97 at 21:14:48
From: Doctor Sydney
Subject: Re: Geometry
Hello Wesley,
Let's first look at a picture that we can refer to when we define
corresponding angles and alternate angles:
/
A / B
----------/-----------
C / D
/
/
/
E / F
-----/------------
G / H
/
There are a lot of letters in this diagram! Don't worry, though,
we'll figure out what everything means in plenty of time.
Assume that the two horizontal lines are parallel. The diagonal is
called a transversal, and as you can see, the intersection of the
transversal with the horizontal lines makes lots of different angles.
In my picture, I gave each of these angles a name: A, B, C, D, E, F,
G, and H.
Whenever you have a set-up like the above in which you have two
parallel lines and a transversal intersecting them, you can think
about corresponding angles and alternate angles.
Two angles are corresponding angles if they are at corresponding spots
in the diagram. Look at the diagram. Do you see how we could easily
split the angles into two groups? A, B, C, and D would be the first
group - they are the angles the transversal makes with the highest
horizontal line. E, F, G, and H would be the second group - they are
the angles the transversal makes with the lowest horizontal line.
Can you see how the bottom set of four angles looks a lot like the top
set of four angles? We say that two angles are corresponding angles
if they occupy corresponding positions in the two sets of angles. For
example, A and E are corresponding angles because they are both in the
"top-left" position: A is in the top left corner of the set of angles
A, B, C, and D, while E is also in the top left corner of the set of
angles E, F, G, and H.
Similarly, C and G are corresponding angles. There are two more pairs
of corresponding angles in the picture. Can you find them?
One neat and helpful fact about corresponding angles is that they are
always equal. Can you see why?
Let's move on to alternate angles. Here is another copy of the
picture I drew above:
/
A / B
----------/-----------
C / D
/
/
/
E / F
-----/------------
G / H
We say that two angles are alternate angles if they fulfill three
requirements:
1. They must be on the "inside" or middle part of the picture. By
inside or middle angles I mean angles C, D, E, and F.
2. They must be on opposite sides of the transversal. Hence A and C
cannot be alternate angles because they are both to the left of the
transversal.
3. If two angles are alternate, one must be from the group of angles
which has the top horizontal line as one of its sides and the other
angle must be from the group of angles which has the bottom
horizontal line as one of its sides. In other words, the last
requirement says that a pair of alternate angles must consist of
one angle from the set {A, B, C, D}, and one angle from the set
{E, F, G, H}.
This sounds complicated, but if we look at the picture and apply the
three requirements, it will become clear what we mean by alternate
angles.
- The first requirement tells us that only C, D, E, and F can
be members of a pair of angles which are parallel. That rules
out a lot of possibilities.
- The second requirement tells us that a pair of alternate angles
must be on opposite sides of the transversal. So, C and E cannot
be a pair of alternate angles. Similarly, D and F cannot be a pair
of alternate angles.
- Applying the final constraint, we see that there are exactly two
pairs of alternate angles in the above picture. One pair is
C and F. C and F fulfill all the requirements of alternate
angles: they are "inside" angles, they are on opposite sides of
the transversal, and they come from different groups of angles.
Can you find the other pair of alternate angles?
A helpful fact about alternate angles is that they, too, are equal.
This fact can make proofs much easier! Can you see why they are
equal?
I hope this helps. Please do write back if you have more questions.
-Doctor Sydney, The Math Forum
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