We have already learned about the criteria SAS, SSS, ASA and HL for congruence of triangles. There is a more general version of each of these tests to shows two triangles are similar.

Suppose we are given two triangles ABC and DEF and wish to check whether or not they are similar. We will use the congruence criteria after scaling triangle ABC suitably.

This will be the scaling step in each case. Let k = |DE|/|AB|. Then on ray CA construct point A' so that |CA'| = k|CA|. Likewise, construct the point B' on ray CB so that |CB'| = k|CB|.

By the Dilation Axiom, the new triangle CA'B' is similar to CAB with scaling constant k, so

• |A'B'| = k|AB|, |CA'| = k|CA|, |CB'| = k|CB|.
• all the angles in CA'B' are equal to the corresponding angle s in CAB.

Each case below will start with triangles ABC and DEF.

1. Triangle A'B'C will be constructed as above.
2. Then it will be shown that triangle A'B'C is congruent to triangle DEF.
3. If this is true, then triangle ABC is similar to A'B'C is similar to DEF with ratio = k*1 = k.

Given triangles ABC and DEF, suppose that angle BAC = angle FDE and for some k > 0, |DE|/|AB| = |DF|/|AC| = k. Then triangle ABC is similar to triangle DEF (with scaling ratio k).

Proof: Construct the scaled triangle A'B'C from ABC as above.

• Because |A'B'|/|AB| = |DE|/|AB| = k then |A'B'| = |DE|. Likewise, |A'C|/|AC| = |DF|/|AC| = k, so |A'C| = |DF|.
• Also, angle B'A'C = angle BAC (similar triangles) and this = angle FDE (given).
• Thus by SAS, triangle A'B'C is congruent to triangle DEF.
• This is the same as saying triangle A'B'C is similar to triangle DEF with scaling ratio 1. Therefore, triangle ABC is similar to A'B'C, which is similar to DEF with ratio = k*1 = k.

Given triangles ABC and DEF, suppose for some k > 0, |DE|/|AB| = |EF|/|BC| = |FD|/|CA| = k. Then triangle ABC is similar to triangle DEF (with scaling ratio k).

Proof: This proof is about the same as the previous one.

• Construct triangle A'B'C as before.
• From |DE|/|AB| = k = |A'B'|/|AB|, if follows that |DE| = |A'B'|. For the same reason |EF| = |B'C| and |FD| = |CA'|.
• Thus by SSS, triangle A'B'C is congruent to triangle DEF.
• The triangle ABC is similar to triangle A'B'C is similar to triangle DEF with ratio k as before.

Given triangles ABC and DEF, suppose angle CAB = angle FDE and angle ABC = angle DEF. Then triangle ABC is similar to triangle DEF. (The scaling ratio is the ratio of any pair of corresponding sides, for example k = |DE|/|AB|).

Proof: This proof again is about the same.

• Construct triangle A'B'C with ratio k = |DE|/|AB|).
• From |DE|/|AB| = k = |A'B'|/|AB|, if follows that |DE| = |A'B'|.
• Also, angle CA'B' = angle CAB (similar triangles) = angle FDE (given) and likewise angle A'B'C = angle ABC = angle DEF.
• Thus by ASA, triangle A'B'C is congruent to triangle DEF.
• The triangle ABC is similar to triangle A'B'C is similar to triangle DEF with ratio k as before.

Note: The ASA criterion for similarity becomes AA, since when only one ratio of sides = k, there is nothing to check.

Given triangles ABC and DEF, suppose angle CAB = angle FDE is a right angle. Then if |EF|/|BC| = |DE|/|AB| = k. Then triangle ABC is similar to triangle DEF (with scaling ratio k).

Proof: This proof follows the same outline as the others. Construct right triangle A'B'C and show it is congruent to DEF by HL. Then ABC is similar to DEF. The details are left to the reader.

The book B&B follows a different path from the Math 444 path to get to the same place.

B&B Principle 5 is the same as SAS for similarity above. In B&B it is given as an axiom. In the Math 444 treatment, the axiom is broken into two parts. SAS for congruence is assumed as an axiom. Then SSS, ASA and HL for congruence are proved as theorems.

On this page, the special case of SAS for similarity called the Dilation Axiom is assumed as a new axiom in the Math 444 path.

• SAS for similarity is proved from the Dilation Axiom and SAS for Congruence. This is the same as B&B Principle 5.
• SSS for similarity is proved above. This is the same as B&B Principle 8.
• AA for similarity is proved above. This is the same as B&B Principle 6
• HL for similarity is proved above. This is the same as B&B Corollary 12a. In the 444 notes, this was proved without Principle 12 (Pythagorean Theorem).

Also, in the Math 444 path, the following B&B Principles have also been proved previously before the Dilation Axiom.

Principle 7: In a triangle, the angles opposite equal side are equal, and conversely. This was proved on the first day in the theory of the isosceles triangle.

Principle10: The set of points equidistant from the endpoint of a segment is the perpendicular bisector of the segment. This was proved several times in class and in homework.

Principle 11: Through a point not on a line there is one and only one perpendicular to the line. This follows from the theory of kites, and is used to define line reflection.

On the other hand, Principles 9 and 12 cannot be proved without the Dilation Axiom.