Similar triangles

How can you measure the height of the flagpole in the school yard on a sunny day without climbing it? Aysel measures the shadow of Elvin, who is 1.6 m tall: it is 2 m. At the same moment the flagpole’s shadow is 15 m. Each shadow is 1.25 times as long as the object itself, so the flagpole is 15 ÷ 1.25 = 12 m tall. Behind this calculation is the main idea of the lesson: similar triangles have the same shape, and their sizes differ by the same factor.

In the lesson “Congruent triangles” we studied triangles with the same shape and the same size, and in the lesson “Quadrilaterals: parallelogram, rectangle, rhombus, square and trapezoid” we met Thales’s theorem. Now we extend Thales’s theorem to proportional segments, learn the three similarity tests and use them for a new proof of “The Pythagorean theorem” and for measuring with shadows. Enlarging a photo and drawing maps and plans rely on similarity too.

Proportional segments and the generalised Thales theorem

In triangle ABC, point M lies on side AB, point N on side BC, and MN ∥ AC. These two parallel lines cross the sides of angle B. By Thales’s theorem, parallel lines that cut equal segments on one side of an angle cut equal segments on the other side too. What if the segments are not equal? Suppose BM : MA = 3 : 2. Divide BM into 3 equal parts and MA into 2, and draw lines parallel to MN through the division points. By Thales’s theorem BC is cut into equal parts as well: 3 of them in BN and 2 in NC, so BN : NC = 3 : 2. The ratio is kept!

Dividing a segment in a given ratio (say, AB in the ratio 2 : 3) is a simple compass-and-ruler construction based on this theorem:

1. 1
   An auxiliary ray

   From A draw any ray that makes an angle with AB.

2. 2
   Equal segments

   With the compass mark off 2 + 3 = 5 equal segments one after another on the ray; call the last point P.

3. 3
   Join

   Join P to B.

4. 4
   Draw a parallel

   Through the second division point draw a line parallel to PB; it meets AB at the required point K: AK : KB = 2 : 3.

The theorem also has a beautiful consequence about the angle bisector of a triangle.

Why? Through C draw a line parallel to BL; it meets the extension of AB beyond B at K. ∠BKC = ∠ABL (corresponding angles) and ∠BCK = ∠LBC (alternate angles), while the bisector makes ∠ABL = ∠LBC. So triangle BCK has two equal angles and is isosceles: BK = BC. Now the parallel lines BL and KC cross the sides of angle A, and by the generalised Thales theorem AL / LC = AB / BK = AB / BC.

Similar triangles and the three similarity tests

As with congruence, the order of the letters shows which vertices correspond: △ABC ∼ △KLM means A ↔ K, B ↔ L, C ↔ M. Corresponding sides lie opposite equal angles. The definition has six conditions, but you do not need to check all of them. The key auxiliary fact is: a line parallel to one side of a triangle cuts off a triangle similar to the given one. In the first drawing, MN ∥ AC gives △MBN ∼ △ABC: angle B is shared, ∠BMN = ∠BAC and ∠BNM = ∠BCA are corresponding angles, and BM / BA = BN / BC comes from the generalised Thales theorem. A line through N parallel to AB gives a parallelogram, and the same theorem shows that MN / AC equals this ratio too.

Why: the third angles are equal as well, because the sum is 180°. Let A₁B₁ > AB. Mark A₁M = AB on A₁B₁ and draw MN ∥ B₁C₁. Triangle A₁MN is similar to the big triangle (parallel line) and congruent to △ABC by ASA. Hence △ABC ∼ △A₁B₁C₁. This is the most used test: two equal angles are often visible in the drawing — a shared angle, vertical angles, angles at parallel lines, right angles.

Both tests are proved the same way: build a “copy” of the small triangle inside the big one (A₁M = AB, MN ∥ B₁C₁) and show that it is congruent to the small triangle with the congruence criteria SAS and SSS.

The ratio of perimeters and of areas

If every side of a similar triangle is k times longer, their sum — the perimeter — is k times longer too. Corresponding altitudes, medians and bisectors are also k times longer, because they are corresponding sides of similar triangles as well. The area, however, is half of base × height: both factors grow k times, so the area grows k · k = k² times.

Proportional segments in a right triangle

In right triangle ABC (∠C = 90°) draw the altitude CH from the right angle. It splits the triangle into two smaller right triangles, and all three are similar to each other: △ACH and △ABC share angle A, △CBH and △ABC share angle B, and each pair also has a right angle (AA). The segments BH = p and AH = q are the projections of the legs a = BC and b = AC onto the hypotenuse.

Where the formulas come from: △CBH ∼ △ACH, so CH / AH = BH / CH, that is h / q = p / h and h² = pq. △CBH ∼ △ABC, so BH / BC = BC / BA, that is a² = cp; in the same way b² = cq. Add the last two: a² + b² = cp + cq = c(p + q) = c². This is a proof of the Pythagorean theorem by similarity! Writing the area in two ways gives one more formula: ½ab = ½ch, so h = ab / c.

Measuring heights with shadows and mirrors

The Sun is so far away that its rays can be treated as parallel. So at a given moment a vertical pole, its shadow and a ray form a right triangle, and a person, their shadow and a ray form a second triangle similar to it: the right angles are equal, and so are the angles the rays make with the ground (AA). The mirror method uses the law of reflection: the angle of incidence equals the angle of reflection, so again we get two similar right triangles.