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How to Construct an Angle Trisection

Instructor: Laura Pennington

Laura received her Master's degree in Pure Mathematics from Michigan State University. She has 15 years of experience teaching collegiate mathematics at various institutions.

Let's discuss what an angle trisection is and how to construct one using just a marked straightedge and a compass. We will also look at a proof solidifying the fact that the construction process does create an angle trisection.

Angle Trisection

Suppose you're studying straightedge and compass constructions in your math class. These types of constructions involve using only an unmarked straightedge and a compass to construct different geometric objects. The teacher says that it was proven in the 19th century that it is impossible to construct an angle trisection using only an unmarked straightedge and a compass. Wait, a what?

An angle trisection is a construction that takes an angle and creates an angle that has a measure of 1/3 of the original angle.


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Since you now know that it is impossible to construct such a trisection using an unmarked straightedge and compass, this gets you to wondering if it is possible to do so in another manner. The answer is absolutely yes, and we can even do it using the two tools mentioned. We just need to use a marked straightedge instead of an unmarked straightedge.

Archimedes' Trisection of an Angle

The process we can use to construct an angle trisection is called Archimedes' trisection of an angle. The steps to this construction are as follows.

Given an angle, ∠ABC, do the following:

1. Use your straightedge and compass to construct a line parallel to line BC that passes through point A.


Step One
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2. Use the compass to construct a circle with center A and radius AB.


Ste Two
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3. On your straightedge, mark the distance between A and B (the radius of the circle), then maneuver it so that we can draw a line segment from B to a point D on the line parallel to BC that intersects the circle at a point E, and DE is the same length as your marked AB.


Step Three
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4. The angle DBC trisects the angle ABC. That is,

  • mDBC = (1/3) mABC

Well, that seems a lot easier than expected! So much so that you may be thinking that you're not convinced! That's okay! Most mathematicians are hard to convince without a proof to back up a claim, so let's take a look at the proof that this construction does, in fact, create an angle trisection.

Proof of Construction

To prove that this construction does, in fact, create a trisection of the angle ABC, we will use the following theorems:

  1. If two parallel lines are cut by a transversal, opposite angles are equal.
  2. If two sides of a triangle have equal length, then the angles opposite those sides have equal measure.
  3. The sum of the angles of a triangle is 180 degrees.

Okay, let's give this proof a go!

Let mDBC = s. Since AD and BC are parallel by construction, theorem 1 above gives that

  • mADB = ∠DBC = s.

Consider triangle AED. Since AE is the radius of the circle we created, it has the same length as DE by our construction. AE = DE, so by theorem 2 above,

  • mEAD = mADB = s.


Picture for First Part of Proof
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By theorem 3 above, we have

  • mEAD + mADB + mAED = s + s + mAED = 2s + mAED = 180

Solving for mAED using the last two parts of this equation gives

  • mAED = 180 - 2s

Since ∠AEB and ∠AED form a straight line, their angles add up to 180, so we have

  • mAEB + ∠AED = mAEB + (180 - 2s) = 180

Solving for mAEB using the last two parts of this equation gives

  • mAEB = 2s

Now consider triangle AEB. Since AE and EB are both radii of the circle, they are equal in length, so by theorem 2 above,

  • mABE = mAEB = 2s.


Picture for Last Part of Proof
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