Book 1 – Proposition 8

Proposition 8

If two triangles have the two sides equal to two sides respectively, and also have the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.

[Desmos graphs can be found here.]

Let ABC and DEF be two triangles such that AB is equal to DE and AC is equal to DF, and that the base BC is equal to the base EF.

We will prove that angle BAC is equal to angle EDF.

If triangle ABC is applied to triangle DEF with the point B placed on the point E and the straight line BC on the straight line EF, then the point C will also coincide with the point F because BC is equal to EF.
BA and AC will also coincide with ED and DF; if they did not then they would fall beside them and coincide with EG and GF for some other point G.

If that were true then given two straight lines (ED and DF) constructed on a straight line EF [from its extremities] and meeting at a point, there has been constructed on the same straight line [from its extremities] and on the same side of it two other straight lines (EG and GF) meeting at another point and equal to ED and DF. Proposition 7 tells us that this construction is not possible.

Therefore BA and AC must coincide with ED and DF. So, the angle BAC will coincide with the angle EDF. Thus the two angles BAC and EDF are equal to one another.

Q.E.D.

In this proof if triangle ABC coincided with triangle DEF, then the two angles BAV and EDF would be equal. So instead we assumed that the triangle ABC did not coincide with triangle DEF and showed that this was not possible.

The next four propositions will consist of constructions. I’d love to hear your feedback on Book 1 Proposition 8, or on anything about this blog – George