Book 1 – Proposition 5

Proposition 5

In isosceles triangles the angles at the base are equal to one another, and, if the equal straight lines be produced further, the angles under the base will be equal to one another.

An isosceles triangle has two sides that are equal. In the picture below the sides AB and AC are equal.

Now we produce the straight lines BD and CE further in a straight line with AB and AC. (Postulate 2)

We are trying to show that the two angles ABC and ACB are equal. Those are the two angles that are opposite the equal sides. We are also trying to prove that angles DBC and ECB are equal. (We will change the labels for those in a moment.)

Place a point F at random on BD, and from AE cut off a straight line AG that is equal to AF (by Proposition 3).

Let the lines FC and GB be joined.

We begin by focusing on triangles ACF and ABG, trying to prove that they are equal.

 Sides AF and AG are equal by our choice of where we placed point G. Side AC is equal to side AB because they are the two equal sides from the isosceles triangle. The pairs of equal sides contain the common angle FAG. So, side CF must be equal to side BG by Proposition 4. Also, the triangle ACF is equal to the triangle ABG, and the angles ACF and ABG are equal, as are angles AFC and AGB.

We now turn our attention to triangles CBF and BCG.

Since AF is equal to AG and AB is equal to AC, the remainder BF is equal to the remainder CG. We proved above that the side CF is equal to the side BG.
So, the two sides BF, FC are equal to the two sides CG, GB respectively, and the angle BFC is equal to the angle CGB. The base BC is common to both triangles. Therefore the triangle CBF is equal to the triangle BCG, and the corresponding angles FCB and GBC are equal. The angles FBC and GCB are also equal to one another.

Now we have all of the tools needed to prove that angles ABC and ACB are equal.

We know that angle ABG is equal to angle ACF, and we know that angle CBG is equal to angle BCF. Therefore, the remaining angle ABC (cutting angle CBG out of angle ABG) and the remaining angle ACB (cutting angle BCF out of angle ACF) must be equal to one another.

We have also proved that the angles under the base will be equal to one another.

When we proved that triangle CBF was equal to the triangle BCG, we showed that the angles FBC and GCB are equal and those are the angles under the base.

Q.E.D.

Here’s a summary of the steps in this proof that angles ABC and ACB are equal.

• Show that triangle ACF is equal to triangle ABG.
  This gives us that angle ACF is equal to angle ABG.
• Show that triangle CBF is equal to triangle BCG.
  This gives us that angle BCF is equal to angle CBG.
  (It also proves that the angles under the base – FBC and GCB – are equal.
• Use the fact that angle ACB remains when angle BCF is cut from angle ACF and that angle ABC remains when angle CBG is cut from angle ABG.

If you want to step through the graphs I used in this proof, you can find them here on Desmos.com.

---

Proposition 6

If in a triangle two angles be equal to one another, the sides which subtend the equal angles will also be equal to one another.

In Proposition 5 we started with two equal sides and proved that the angles that were opposite of those sides were equal. In this proposition we will begin with two equal angles and prove that the two sides that are opposite those angles are equal.

This proof will be interesting. We will assume that the two sides are not equal and reach a contradiction that proves our assumption must be false.

If you have any questions or feedback on the proof of Proposition 5, please leave a comment – George