Proving and Applying Similar Triangles
Theory
Two triangles are similar when their angles match, so their sides are in the same ratio. This Year 10A maths page covers proving that triangles are similar with the AA, SAS, SSS and RHS tests, writing statements in matching order, nested and hourglass triangles, equations with similar triangles and real problems such as shadows.
Similar triangles have equal matching angles and matching sides in the same ratio \(k\). The symbol is \(\sim\), with the vertices in matching order.
Tests: AA, two pairs of equal angles (the third pair then matches too); SAS, two pairs of sides in the same ratio and the included angles equal; SSS, all three pairs of sides in the same ratio; RHS, right angles with hypotenuses and one other pair of sides in the same ratio.
The equal angles usually come from a common angle, vertically opposite angles, parallel lines (corresponding or alternate) or two right angles.
Once similarity is proved, write \(\dfrac{a}{a'} = \dfrac{b}{b'} = \dfrac{c}{c'}\) for matching sides and solve for the unknown.
| Shape | Equal angles from | Watch for |
|---|---|---|
| nested, parallel side | common angle, corresponding angles | use whole sides \(AB\), not the piece \(DB\) |
| nested, not parallel | common angle, one given pair | the small triangle is flipped: match by angles |
| hourglass | vertically opposite, alternate angles | the second triangle is upside down |
| right triangle with altitude | common angle, right angles | three similar triangles |
| shadows, mirrors, rebounds | right angles, equal angles of light | same units on both triangles |
How to solve a similar-triangles problem
- Find two triangles; redraw them separately, the same way up, if they overlap.
- Prove similarity: give two pairs of equal angles with reasons (or the side ratios for SAS or SSS).
- Write the statement in matching order, then the ratios of matching sides in that order.
- Substitute and solve; cross-multiply when the unknown is in a denominator.
- Check that the answer is sensible: the larger triangle's sides must be the larger ones.
Practical and harder problems
- Heights and widths: shadows, pinhole cameras, river widths and ramps all make two right-angled triangles sharing an angle.
- Unknown in both triangles: write the larger triangle's side as a sum, such as \(s + 3\) or \(120 - x\), then solve (Example 4).
- Pythagoras first: sometimes a missing side, such as a ladder's height up a wall, must be found before the ratio can be used.
- Shapes inside triangles: a square or rectangle with corners on the sides creates a small triangle similar to the whole one; write its sides in terms of the unknown.
- Proving a product of lengths: to show a result such as \(PQ^2 = PS \times PR\), write the ratio of matching sides from the similar triangles, then cross-multiply.
- Areas: find the matching lengths with the scale factor first, then work out the area of the new triangle.
- Two crossing wires between poles: each wire makes a pair of similar triangles; write the crossing height two ways and equate them.
(i) Prove that \(\triangle ADE \sim \triangle ABC\).
(ii) Find \(BC\) and \(EC\).
(i)
\(\angle A \text{ is common}\)
\(\angle ADE = \angle ABC\) (corresponding angles, \(DE \parallel BC\))
\(\therefore \triangle ADE \sim \triangle ABC\) (AA)
(ii) Use whole sides of the big triangle: \(AB = 4 + 6 = 10\), so \(k = \dfrac{10}{4} = 2.5\).
| \(BC\) | \(=\) | \(5 \times 2.5 = \mathbf{12.5}\) |
| \(AC\) | \(=\) | \(3.2 \times 2.5 = 8\) |
| \(EC\) | \(=\) | \(8 - 3.2 = \mathbf{4.8}\) |
Using \(\dfrac{DB}{AD} = \dfrac{6}{4}\) as the scale factor would be wrong: \(DB\) is not a side of either triangle.
(i) Write a similarity statement.
(ii) Find \(PQ\) and \(QR\).
(i) \(\angle P\) is common and \(\angle PST = \angle PRQ\). Match the vertices by angles, not by position:
| \(P \leftrightarrow P,\ S \leftrightarrow R,\ T\) | \(\leftrightarrow\) | \(Q\) |
| \(\triangle PST\) | \(\sim\) | \(\triangle PRQ \quad (\text{AA})\) |
(ii) Write the three ratios in the same order as the statement.
| \(\dfrac{PS}{PR} = \dfrac{PT}{PQ}\) | \(=\) | \(\dfrac{ST}{RQ}\) |
| \(\dfrac{3}{9} = \dfrac{4}{PQ}\) | \(=\) | \(\dfrac{6}{RQ}\) |
| \(PQ\) | \(=\) | \(4 \times 3 = \mathbf{12}\) |
| \(QR\) | \(=\) | \(6 \times 3 = \mathbf{18}\) |
\(PS\) matches \(PR\), not \(PQ\). Writing the triangles in matching order first prevents this mistake.
\(\angle BAE = \angle DCE\) (alternate angles, \(AB \parallel DC\)) and \(\angle AEB = \angle CED\) (vertically opposite), so \(\triangle ABE \sim \triangle CDE\) (AA).
| \(\dfrac{AE}{CE}\) | \(=\) | \(\dfrac{BE}{DE}\) |
| \(\dfrac{x + 1}{6}\) | \(=\) | \(\dfrac{4}{x + 3}\) |
| \((x + 1)(x + 3)\) | \(=\) | \(24\) |
| \(x^2 + 4x - 21\) | \(=\) | \(0\) |
| \((x + 7)(x - 3)\) | \(=\) | \(0\) |
\(x = -7\) gives a negative length, so \(x = \mathbf{3}\). Check: \(\dfrac{4}{6} = \dfrac{4}{6}\).
The small triangle (person and shadow) and the large triangle (lamp post and the ground to the shadow tip) share the angle at the tip and both have a right angle, so they are similar (AA).
Let the shadow be \(s\) m. The base of the large triangle is \(s + 3\).
| \(\dfrac{s}{s + 3}\) | \(=\) | \(\dfrac{1.8}{4.5}\) |
| \(4.5s\) | \(=\) | \(1.8s + 5.4\) |
| \(2.7s\) | \(=\) | \(5.4\) |
| \(s\) | \(=\) | \(\mathbf{2}\text{ m}\) |
Check: \(\dfrac{2}{5} = 0.4 = \dfrac{1.8}{4.5}\). Whenever the unknown appears in both triangles, write each side in terms of it and cross-multiply.
Common pitfalls
Frequently asked questions
How do you prove two triangles are similar?
Show two pairs of equal angles (AA), or three pairs of sides in the same ratio (SSS), or two pairs of sides in the same ratio with the included angles equal (SAS), or the RHS version with ratios. Give a reason for each fact.
What is the difference between similar and congruent triangles?
Congruent triangles are the same shape and the same size. Similar triangles are the same shape but can be different sizes, with sides in a fixed ratio.
How do you find a missing side in similar triangles?
Write the triangles in matching order, set up the ratios of matching sides, substitute the known lengths and solve for the unknown, cross-multiplying if needed.
Why are the triangles in an hourglass shape similar?
The angles where the lines cross are vertically opposite, so they are equal, and the parallel top and bottom give a pair of equal alternate angles. Two pairs of equal angles prove similarity.
How do you use similar triangles to find the height of a tree?
At the same time of day, a stick and the tree make shadows with the same angle of the sun. The heights are in the same ratio as the shadow lengths, so tree height equals stick height times tree shadow divided by stick shadow.