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🏰 Geometry Kingdom · Geometry

Similar Triangles

Recognise similar figures, write proportions between corresponding sides, and use them to find unknown lengths.

In short

  • Similar triangles have equal angles and sides in a constant ratio — the scale factor.
  • The letter order in ABC ~ DEF tells you which sides correspond; write it down first.
  • Two equal angles (AA) are enough to prove similarity, which is what makes shadow measurement work.
  • Lengths scale by k, areas by k2 and volumes by k3.
  • A line parallel to one side of a triangle cuts the other two sides in the same ratio, and the segment joining two midpoints is parallel to the third side and half as long.

Same shape, different size

Two triangles are similar when one is an enlargement of the other: the angles match exactly and every side has been multiplied by the same number, the scale factor.

The symbol is a tilde: ABC ~ DEF. The order of the letters carries information — A pairs with D, B with E, C with F. That means AB pairs with DE, BC with EF, and AC with DF.

Writing the correspondence down before doing any arithmetic prevents most of the errors in this topic.

Similarity is about multiplying, never about adding. If one triangle has sides 3, 4, 5 and another has sides 6, 8, 10, the scale factor is 2 — not "plus 3, plus 4, plus 5".

How to know two triangles are similar

You do not need to check everything.

  • AA: if two angles of one triangle equal two angles of the other, the triangles are similar. (The third angle is then forced by the 180° angle sum.) This is by far the most used test.
  • SSS: if all three pairs of matching sides are in the same ratio, the triangles are similar.
  • SAS: if two pairs of sides are in the same ratio and the angles between them are equal, the triangles are similar.

AA is what makes shadow and mirror problems work: the sun's rays arrive at the same angle for everything nearby, and everything stands vertically on level ground, so the two triangles have the same angles.

Finding a missing side

Once similarity is established, matching sides are in proportion, and a missing length comes from a single equation.

Suppose ABC ~ DEF with AB = 4, BC = 6 and DE = 10. Then

DE / AB = EF / BC 10 / 4 = EF / 6

The scale factor is 10/4 = 2.5, so EF = 2.5 x 6 = 15.

Two ways to go wrong, and both are easy to catch. First, check that the answer moves in the right direction — enlarging should make the number bigger. Second, verify with a second pair of sides if you have one.

Real measuring uses exactly this. A 1.6 m person casting a 2 m shadow beside a tower casting a 25 m shadow gives a scale factor of 25/2 = 12.5, so the tower is 12.5 x 1.6 = 20 m tall.

How perimeter and area scale

This is the part that surprises people. If the scale factor for lengths is k, then

  • every length (side, perimeter, height, diagonal) is multiplied by k
  • every area is multiplied by k2

A perimeter is just a sum of lengths, so it scales like a length. An area is a length times a length, so it picks up the factor twice.

Doubling a triangle (k = 2) doubles its perimeter but makes its area four times as big. Tripling it multiplies the area by 9.

The same rule extends to solids: volumes scale by k3, which is why a model that is half scale holds only an eighth as much.

Cutting one triangle into two

Similarity does not need two separate triangles. Draw a line across one triangle and you often get a smaller triangle sitting inside the original, sharing an angle with it — and three theorems come out of that.

The side-splitter theorem. In triangle ABC, let D be a point on AB and E a point on AC. If DE is parallel to BC, then the two sides are cut in the same ratio:

AD / DB = AE / EC

The reason is AA: DE parallel to BC makes angle ADE equal to angle ABC, angle A is shared, so triangle ADE is similar to triangle ABC. The theorem works backwards too — if the ratios are equal then DE must be parallel to BC, which is how "is DE parallel to BC?" gets answered: work out both fractions and see whether they agree.

The trap is the whole standing in for a part. DB is the piece from D to B, not the whole of AB. If a question gives you AB, subtract AD first.

The midsegment theorem. Now let D and E be the *midpoints* of AB and AC. The segment DE is called a midsegment, and it is parallel to BC and exactly half as long:

DE = BC / 2

So a midsegment of a triangle with BC = 18 cm is 9 cm, and going the other way, a midsegment of 7 cm means BC is 14 cm. Every side of triangle ADE is half the matching side of ABC, so its perimeter is half the perimeter of ABC as well.

The angle-bisector theorem. If the bisector of angle A meets BC at D, then D cuts BC in the ratio of the two sides that meet at A:

BD / DC = AB / AC

The longer of AB and AC always sits beside the longer piece of BC, which is a quick way to check an answer.

Lengths in this topic are answered in centimetres as a whole number, unless the question asks for one decimal place — and then it says so.

Worked examples

Example 1

ABC ~ DEF. AB = 5 cm, BC = 8 cm and DE = 15 cm. Find EF.

  1. Match the letters: AB pairs with DE, and BC pairs with EF.
  2. Scale factor from ABC to DEF: 15 / 5 = 3.
  3. Every side of DEF is 3 times the matching side of ABC.
  4. EF = 3 x 8 = 24 cm.

Example 2

Two similar triangles have scale factor 4. The smaller has area 7 cm2. Find the area of the larger.

  1. Lengths are multiplied by 4, so both dimensions of the shape grow by 4.
  2. Area therefore grows by 42 = 16.
  3. Area = 7 x 16.
  4. The larger triangle has area 112 cm2.

Example 3

In triangle ABC, D lies on AB and E lies on AC, and DE is parallel to BC. AB = 20 cm, AD = 8 cm and AE = 6 cm. Find EC.

  1. DE is parallel to BC, so the side-splitter theorem gives AD / DB = AE / EC.
  2. DB is only the upper part of AB, so DB = AB - AD = 20 - 8 = 12 cm.
  3. Substitute what is known: 8 / 12 = 6 / EC.
  4. Cross-multiply: 8 x EC = 12 x 6 = 72, so EC = 72 / 8.
  5. EC = 9 cm. Check: 8 / 12 and 6 / 9 both simplify to 2 / 3.

Example 4

In triangle ABC, the bisector of angle A meets BC at D. AB = 12 cm, AC = 18 cm and BD = 6 cm. How long is BC?

  1. The angle-bisector theorem gives BD / DC = AB / AC.
  2. Substitute: 6 / DC = 12 / 18.
  3. Cross-multiply: 12 x DC = 6 x 18 = 108, so DC = 9 cm.
  4. AC is the longer of the two sides at A, and DC came out as the longer piece, so the pairing is the right way round.
  5. BC is the whole of the side: BC = BD + DC = 6 + 9 = 15 cm.

Practice problems, with solutions

Three problems of increasing difficulty, each with the full working. In the game these are generated fresh every time; these three are fixed so this page always shows the same ones.

Problem 1

Difficulty 1 of 5

Triangle ABC has sides 6, 7 and 9. Triangle DEF is similar to it, with matching sides 18, 21 and 27. What is the scale factor from ABC to DEF?

Answer: 3

  1. Scale factor = (side of DEF) / (matching side of ABC)
  2. k = 18 / 6 = 3
  3. Check: 21 / 7 = 3 as well.

Problem 2

Difficulty 3 of 5

Triangle ABC is similar to triangle DEF (ABC ~ DEF). AB = 12 cm, BC = 2 cm and DE = 30 cm. How long is EF?

Answer: 5 cm

  1. ABC ~ DEF, so DE / AB = EF / BC.
  2. Scale factor k = 30 / 12 = 2.5
  3. EF = k x BC = 2.5 x 2
  4. EF = 5 cm

Problem 3

Difficulty 4 of 5

At the same moment, Zara (1.7 m tall) casts a shadow 1.5 m long, and a flagpole casts a shadow 10.5 m long. How tall is the flagpole?

Answer: 11.9 m

  1. The two triangles are similar, so height / base matches: 1.7 / 1.5 = h / 10.5.
  2. Scale factor = 10.5 / 1.5 = 7
  3. h = 7 x 1.7 = 11.9 m

Common mistakes

  • Adding a constant difference to each side instead of multiplying by a scale factor.
  • Pairing the wrong sides because the correspondence was never written down.
  • Turning the ratio upside down, so the answer shrinks when it should grow.
  • Scaling an area by k instead of by k2.
  • Putting the whole side where a part belongs. In AD / DB = AE / EC, DB is the piece from D to B, so subtract AD from AB before dividing.
  • Treating a midsegment as the same length as the third side. It is parallel to that side but only half of it, so halving or doubling is always part of the answer.
  • Turning the angle-bisector ratio upside down. BD sits beside AB and DC beside AC, so the longer side at A always meets the longer piece of the opposite side.

What you should be able to do

  • Identify similar triangles from angle relationships.
  • Write and solve a proportion between corresponding sides, including the side-splitter, midsegment and angle-bisector theorems.
  • Use a scale factor to find perimeters and areas.
  • Solve indirect measurement problems such as shadow heights.

Where this fits in the curriculum

Common Core

  • 7.G.A.1

    Grade 7 — Solve problems involving scale drawings, including computing lengths from a scale.

  • 8.G.A.4

    Grade 8 — Understand similarity through dilations, and describe a sequence that shows two figures are similar.

  • HSG-SRT.B.5

    High school — Use congruence and similarity criteria for triangles to solve problems and prove relationships.

  • HSG-SRT.B.4

    High school — Prove theorems about triangles, including that a line parallel to one side divides the other two proportionally, and prove the Pythagorean theorem using triangle similarity.

Ontario

  • G7.E1.3

    Grade 7 — Perform dilations and describe the similarity between the image and the original shape.

  • G8.E1.3

    Grade 8 — Use scale drawings to calculate actual lengths and areas, and reproduce a drawing at a different ratio.

    Ontario has no similar-triangle criteria (AA, SSS, SAS) before Grade 10: dilations and scale drawings are as close as the elementary curriculum comes, and the formal treatment waits for MPM2D.

SAT

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