Proportions in similar triangles
Two triangles are known to have the same shape. How do the sides you can measure give you the one you cannot?
Where you are 10 in Similarity
When you reach for it
Two triangles are known to be similar, most often because two pairs of their angles are congruent. Some of their sides have been measured and one has not. Every side of one triangle is the same multiple of its partner in the other, so the measured sides give the missing one through a Proportion.
The move turns up in two kinds of problem, one per day. On the first day the triangles are drawn on the page: side by side, turned or flipped, or one inside the other. On the second they are in the world, and the missing side is a length nobody can put a tape measure on, such as the height of a lamp or the width of a river. Finding a length from other lengths, without measuring it, is called indirect measurement. Every method of it finds a small triangle that can be measured completely and a large triangle with the same angles that contains the length wanted.
The procedure
The matching comes from the angles
A similarity statement such as \triangle PQR \sim \triangle STU is a list of matched vertices: P with S, Q with T and R with U. A side is named by its two ends, so the matched vertices fix the matched sides: PQ with ST, QR with TU, and RP with US. Every pair is in one ratio, the scale factor k:
When the triangles are drawn with arcs, the vertices carrying the same arcs are matched, wherever they sit on the page. Said another way, matching sides sit opposite matching angles.
Writing and checking the proportion
Take one ratio whose two lengths are both known and one that contains the unknown, and set them equal. Suppose \triangle PQR \sim \triangle STU with PQ = 8, QR = 14 and ST = 12, and TU is wanted. TU matches QR, so
Multiplying both sides by 14 keeps the equation true (the multiplication property of equality) and gives \displaystyle TU = \frac{12 \cdot 14}{8} = 21. Cross-multiplying is the same step written another way; solving a proportion covers it in full.
What makes the proportion right is the order: the second triangle’s side on top in both ratios, or underneath in both. Mixing them sets k equal to \displaystyle \frac{1}{k}, which holds only when the triangles are the same size. There is a second correct form. Multiplying both sides of \displaystyle \frac{ST}{PQ} = \frac{TU}{QR} by \displaystyle \frac{PQ}{TU} gives \displaystyle \frac{ST}{TU} = \frac{PQ}{QR}, which compares two sides of the same triangle: the ratio of two sides belongs to the shape, not the size. Either form gives the same answer.
To check, divide the answer by its partner. The result must be the scale factor from the known pair: \displaystyle \frac{12}{8} = 1.5, and \displaystyle \frac{21}{14} = 1.5 as well.
Nested triangles
Often a segment cuts across a triangle and cuts off a smaller triangle at one corner. The angle at that corner belongs to both triangles, which is one pair of congruent angles. If the segment is parallel to the opposite side, the angles it makes are Corresponding angles with the angles of the large triangle, and the matching goes corner to nearby corner. If instead the segment makes the other angle equal, the matching crosses over, as Example 3 shows.
The safest way to read nested triangles is to pull them apart and copy the lengths onto each. Then one fact is plain: the side of the large triangle that matches a piece of the small one is the whole side, the piece plus the rest. The rest on its own is not a side of either triangle. (The triangle proportionality theorem is a different result, about the pieces.)
Where the angles come from in indirect measurement
Each method produces two pairs of congruent angles without measuring any angle. Two of the methods rest on a fact about light, taken as an assumption: the sun is so far away that its rays are treated as parallel, and a flat mirror sends light away at the same angle to it as the light arrived.
| Method | First pair of angles | Second pair of angles |
|---|---|---|
| a shadow in sunlight | right angles: upright objects on level ground | parallel rays meet the ground at the same angle |
| a mirror on the floor | right angles at the feet | equal angles at the mirror |
| a shadow from a lamp | right angles at both feet | the angle at the tip of the shadow is shared |
| stakes across a river | right angles laid out at two stakes | Vertical angles where two lines cross |
Naming the two pairs is the reason the proportion is allowed, and it shows which length matches which. Shadows in sunlight and the mirror are taught in full in measuring what you cannot reach. The examples here take the lamp, where the two triangles overlap, and the stakes, where they cross.
Worked examples
Examples 1 to 3 belong to the first day: drawn triangles, including nested ones. Examples 4 and 5 belong to the second: indirect measurement.
A triangle turned and flipped
In the figure, \angle A \cong \angle Y (one arc) and \angle B \cong \angle Z (two arcs). BC = 10, YZ = 12, ZX = 16 and XY = 14. Find AB = x and CA = y.
Two pairs of congruent angles make the triangles similar by AA. A goes with Y and B with Z, so the third vertices go together, C with X:
Reading the sides from the letters: AB matches YZ, BC matches ZX, and CA matches XY. The same follows from opposite angles: BC is opposite \angle A and ZX is opposite \angle Y, the two one-arc angles. The large triangle is turned and mirrored, so its sides cannot be paired by where they sit.
The pair with both lengths known is BC and ZX. Small triangle on top in every ratio:
Multiplying both sides by 12 gives \displaystyle x = \frac{120}{16} = 7.5. In the same way
so \displaystyle y = \frac{140}{16} = 8.75.
Check with the scale factor from the small triangle to the large one, \displaystyle k = \frac{16}{10} = 1.6: \displaystyle \frac{12}{7.5} = 1.6 and \displaystyle \frac{14}{8.75} = 1.6.
A triangle inside a triangle
In \triangle PQR, S lies on PQ and T lies on PR, with ST \parallel QR. PS = 9, ST = 6 and QR = 10. Find SQ.
First the angles. The angle at P belongs to both \triangle PST and \triangle PQR. Line PQ crosses the parallel lines ST and QR, and \angle PST and \angle PQR sit in matching positions at the two crossings, so they are congruent corresponding angles. By AA
with S matching Q and T matching R. Pulled apart, as on the right of the figure, the small triangle has sides PS = 9 and ST = 6, and the large one has the matching sides PQ and QR = 10. Write SQ = x. The side PQ is the whole side, PS + SQ = 9 + x.
Cross-multiplying gives 90 = 6(9 + x). The distributive property turns the right side into 54 + 6x, so 6x = 36 and x = 6. So SQ = 6.
Check: PQ = 15, and \displaystyle \frac{15}{9} = \frac{5}{3}, while \displaystyle \frac{10}{6} = \frac{5}{3}.
The proportion pairs PS with all of PQ, never with the piece SQ. Since SQ is not a side of either triangle, nothing makes \displaystyle \frac{9}{x} equal to \displaystyle \frac{6}{10}; that equation would give x = 15, which fails the check.
When the inside segment is not parallel
In \triangle ABC, AB = 16 and BC = 14. E lies on AC with AE = 6 and EC = 6, and D lies on AB with \angle AED \cong \angle ABC. Find AD, DE and DB.
The angle at A is shared, and \angle AED \cong \angle ABC is given. By AA the triangles are similar, and the given angles decide the matching: E goes with B. The shared vertex A goes with itself, so the remaining vertices D and C go together.
It is not \triangle ADE \sim \triangle ABC, even though D sits on side AB. Reading the letters: AE matches AB, ED matches BC, and DA matches CA. The side CA is the whole side, 6 + 6 = 12.
The pair with both lengths known is AE and AB, so the scale factor from the large triangle to the small one is
Every side of the small triangle is \displaystyle \frac{3}{8} of its partner: \displaystyle AD = \frac{3}{8} \cdot 12 = 4.5 and \displaystyle DE = \frac{3}{8} \cdot 14 = 5.25. Then, by segment addition, DB = AB - AD = 16 - 4.5 = 11.5.
The answer confirms the picture. Had DE been parallel to BC, D would split AB in the same ratio as E splits AC, half way, at AD = 8. Instead AD = 4.5, so DE is not parallel and the crossed matching was the right one.
A lamp post at night
At night Heloise, who is 1.65 m tall, stands 3 m from the foot of a lamp post. The lamp at its top throws her shadow straight away from the post, and the shadow is 2.5 m long. How high is the lamp?
The ray from the lamp that just clears the top of Heloise’s head lands at the tip of her shadow. So there are two triangles: a small one made by Heloise and her shadow, and a large one made by the post and the ground from its foot to the tip of the shadow, with the same ray along the top of both.
A lamp’s rays are not parallel; they spread out from one point. The angles come from elsewhere:
- The post and Heloise both stand upright on level ground: a right angle at each foot.
- The angle at the tip of the shadow is one angle, and it belongs to both triangles.
By AA the small triangle is similar to the large one, with Heloise matching the post and her shadow matching the whole ground side of the large triangle. That side is 3 + 2.5 = 5.5 m, not 3 m. Call the lamp’s height h:
so h = 1.65 \cdot 2.2 = 3.63. The lamp is 3.63 m above the ground.
Check with the ratio inside each triangle: \displaystyle \frac{1.65}{2.5} = 0.66 and \displaystyle \frac{3.63}{5.5} = 0.66.
The width of a river
Jolanta wants the width of a river without crossing it. She picks a tree T on the far bank and puts a stake at B on the near bank, directly opposite the tree, so that TB is at right angles to the bank. She walks 24 m along the bank and puts a stake at C, then 8 m further and puts a stake at D. At D she turns at right angles, away from the river, and walks until stake C exactly hides the tree. There, at E, she has walked 9 m. How wide is the river?
The sight line from E through C to T is straight, and it crosses the bank line BD at C. The two pairs of congruent angles:
- \angle TBC and \angle EDC are both right angles; Jolanta laid them out that way.
- \angle TCB and \angle ECD are Vertical angles, made by two straight lines crossing at C.
By AA
with T matching E, B matching D and C matching itself. So TB matches ED, and BC matches DC. Call the width w:
so w = 3 \cdot 9 = 27. The river is 27 m wide.
Check: the scale factor is \displaystyle \frac{24}{8} = 3, and \displaystyle \frac{27}{9} = 3. Every length Jolanta measured was on dry land.
Boundaries
The triangles must be similar first. A proportion between sides is allowed only once two pairs of congruent angles (or one of the side tests in SSS and SAS similarity) have shown the triangles are similar. Two triangles that merely look alike give a number that means nothing.
Matching sides, in the same order. The proportion is between sides that match under the similarity statement, with the same triangle on top in both ratios. The matching comes from the angles, never from how the triangles are drawn.
The whole side in nested triangles. The large triangle’s side is the full length from the shared vertex, piece plus rest. A piece on its own belongs to no triangle in the figure.
Lengths only. The scale factor multiplies lengths: sides, heights, perimeters. Angles do not change at all, and areas scale by k^2, not k (scale factor and perimeter, area and volume).
The assumptions of the scene. Indirect measurement is only as good as the facts that made the angles congruent: the objects upright, the ground level, sun shadows measured at the same moment, the mirror flat, the stakes’ right angles true. The answer is also only as accurate as the measurements that went in.
A lamp is not the sun. Under a lamp the rays spread out from one point, so no single height-to-shadow ratio holds for everyone; the only similar triangles are the ones sharing a ray, as in Example 4.
Where this goes next
The second form of the proportion, a ratio of two sides inside one triangle, is the start of trigonometry. Every right triangle with a given acute angle is similar to every other, so the ratio of two of its sides depends on the angle alone; those ratios are the sine, cosine and tangent, and right triangle trigonometry uses them to replace the second triangle altogether. Sightings from two places are taken up in angles of elevation and depression.
Nested triangles with a parallel side lead to the triangle proportionality theorem and its proof in proving the triangle proportionality theorem. The three similar triangles inside a right triangle give geometric means in a right triangle. Writing out why two triangles are similar as a full argument is proofs about similar figures, and choosing what goes on top of each ratio when the story is in words is setting up proportions from word problems.
Exercises
Exercises 1 to 11 go with the first day (Examples 1 to 3), Exercises 12 to 17 with the second (Examples 4 and 5), and the reasoning group draws on both.
Day 1: matching sides
- \triangle ABC \sim \triangle DEF. Name the three pairs of matching sides, and write a proportion that links AB, BC, DE and EF.
- \triangle JKL \sim \triangle MNO, with JK = 10, KL = 6, LJ = 12 and MN = 15. Find NO and OM.
- \triangle ABC \sim \triangle XYZ, with AB = 7, BC = 11 and XY = 5. Find YZ as a fraction.
- In \triangle GHI and \triangle PQR, \angle G \cong \angle R and \angle H \cong \angle P. GH = 6, HI = 9, GI = 12 and RP = 10. Find PQ and RQ.
- \triangle ABC \sim \triangle DEF, with AB = 4, DE = 10, BC = x + 1 and EF = 3x - 2. Find x, BC and EF.
- Ishaan claims that \triangle PQR \sim \triangle XYZ, where PQ = 6, QR = 10, XY = 9 and YZ = 14. Use the scale factor to show that the claim cannot be right.
Day 1: nested triangles
In Exercises 7 to 10, D lies on side AB of \triangle ABC, E lies on side AC, and DE \parallel BC.
- AD = 4, AB = 10 and DE = 3. Find BC.
- AE = 5, EC = 7 and BC = 18. Find DE.
- AD = 8, DE = 5 and BC = 12. Find DB.
- AD = x, DB = 9, DE = 4 and BC = 10. Find x.
- In \triangle PQR, S lies on PQ and T lies on PR, with \angle PTS \cong \angle PQR. PQ = 15, PR = 10, QR = 12 and PS = 4. Write the similarity statement, then find PT and ST.
Day 2: shadows and mirrors
- Ishaan is 1.62 m tall and casts a shadow 1.35 m long. At the same moment a clock tower on the same level ground casts a shadow 7.5 m long. Name the two pairs of congruent angles, then find the height of the tower.
- Jolanta, who is 1.72 m tall, stands 4 m from the foot of a lamp post at night, and her shadow, pointing straight away from the post, is 2 m long. How high is the lamp?
- A street lamp is 5.4 m above level ground. Kaspar, who is 1.8 m tall, stands 6 m from the foot of its post. How long is his shadow?
- A column in a museum courtyard is 9 m tall. A mirror lies flat on the level ground 12 m from its foot. Kaspar’s eyes are 1.65 m above the ground. How far from the mirror must he stand, in line with the column, to see the top of the column in it?
Day 2: stakes
- Faustina measures a river as Jolanta did in Example 5. Her stakes give BC = 30 m and CD = 12 m, and the sight line through C reaches the tree when DE = 7 m. Name the two pairs of congruent angles, and find the width of the river.
- Ishaan stands at A on the near rim of a canyon, directly across from a rock R on the far rim. He walks 20 m straight back from the rim, along line RA, to B. At A he marks a point D that is 12 m from A, with AD at right angles to RA; at B he marks a point C on the same side, 18 m from B, with BC at right angles to RB, placed so that R, D and C lie on one line. How wide is the canyon from R to A?
Reasoning
- In Example 2, explain why \displaystyle \frac{PS}{SQ} = \frac{ST}{QR} is not a correct proportion, and use the numbers of that example to show that it fails.
- Two people of the same height stand at different distances from the same street lamp. Explain why their shadows have different lengths, although under the sun their shadows would be the same length.
- In the river measurement of Example 5, suppose Jolanta walks the same distance from C to D as from B to C. Explain why DE is then exactly the width of the river.
Answers
- AB matches DE, BC matches EF and CA matches FD. One proportion is \displaystyle \frac{AB}{DE} = \frac{BC}{EF}; \displaystyle \frac{AB}{BC} = \frac{DE}{EF} is also correct.
- NO = 9 and OM = 18. The scale factor is \displaystyle k = \frac{MN}{JK} = \frac{15}{10} = 1.5. NO matches KL, so NO = 1.5 \cdot 6 = 9; OM matches LJ, so OM = 1.5 \cdot 12 = 18.
- \displaystyle YZ = \frac{55}{7}. YZ matches BC, so \displaystyle \frac{YZ}{11} = \frac{5}{7} and \displaystyle YZ = \frac{5 \cdot 11}{7}.
- PQ = 15 and RQ = 20. The angles give G with R and H with P, so I goes with Q: \triangle GHI \sim \triangle RPQ. Then GH matches RP, HI matches PQ and GI matches RQ. With \displaystyle k = \frac{10}{6} = \frac{5}{3}, \displaystyle PQ = \frac{5}{3} \cdot 9 = 15 and \displaystyle RQ = \frac{5}{3} \cdot 12 = 20.
- x = 9, BC = 10 and EF = 25. \displaystyle \frac{3x - 2}{x + 1} = \frac{10}{4}, so 4(3x - 2) = 10(x + 1), which is 12x - 8 = 10x + 10, and x = 9. Check: \displaystyle \frac{25}{10} = 2.5 = \frac{10}{4}.
- With this matching, \displaystyle \frac{XY}{PQ} = \frac{9}{6} = 1.5 but \displaystyle \frac{YZ}{QR} = \frac{14}{10} = 1.4. Similar triangles have one scale factor for every pair of matching sides, so \triangle PQR \sim \triangle XYZ is false.
- BC = 7.5. The shared angle at A and the corresponding angles \angle ADE \cong \angle ABC give \triangle ADE \sim \triangle ABC, so \displaystyle \frac{DE}{BC} = \frac{AD}{AB}: \displaystyle \frac{3}{BC} = \frac{4}{10} and BC = 7.5.
- DE = 7.5. The whole side is AC = 5 + 7 = 12, so \displaystyle \frac{DE}{18} = \frac{5}{12} and \displaystyle DE = \frac{90}{12} = 7.5.
- DB = 11.2. \displaystyle \frac{AB}{8} = \frac{12}{5}, so the whole side is AB = 19.2, and DB = 19.2 - 8 = 11.2.
- x = 6. The whole side is AB = x + 9, so \displaystyle \frac{x}{x + 9} = \frac{4}{10}. Cross-multiplying, 10x = 4x + 36, so 6x = 36.
- \triangle PTS \sim \triangle PQR, PT = 6 and ST = 4.8. The angle at P is shared and \angle PTS \cong \angle PQR, so T goes with Q and S with R. Then PS matches PR, so \displaystyle k = \frac{4}{10} = 0.4; PT = 0.4 \cdot 15 = 6 and TS = 0.4 \cdot 12 = 4.8.
- 9 m. Ishaan and the tower both stand upright on level ground, a pair of right angles; the sun’s rays are parallel, so they make congruent corresponding angles with the ground. Then \displaystyle \frac{h}{1.62} = \frac{7.5}{1.35}, so \displaystyle h = \frac{1.62 \cdot 7.5}{1.35} = 9.
- 5.16 m. The large triangle’s ground side runs from the post to the tip of the shadow, 4 + 2 = 6 m. With right angles at both feet and the shared angle at the shadow’s tip, \displaystyle \frac{h}{1.72} = \frac{6}{2}, so h = 3 \cdot 1.72 = 5.16.
- 3 m. Call the shadow s. The large triangle’s ground side is 6 + s, so \displaystyle \frac{s}{6 + s} = \frac{1.8}{5.4} = \frac{1}{3}. Then 3s = 6 + s, so s = 3.
- 2.2 m. Right angles at both feet and equal angles at the mirror make the triangles similar. Call the distance d. His eye height matches the column and d matches 12, so \displaystyle \frac{d}{12} = \frac{1.65}{9} and \displaystyle d = \frac{12 \cdot 1.65}{9} = 2.2.
- 17.5 m. The angles at B and D are right angles, and the angles at C are vertical angles, so \triangle TBC \sim \triangle EDC. Then \displaystyle \frac{w}{7} = \frac{30}{12}, so w = 2.5 \cdot 7 = 17.5.
- 40 m. The angle at R is shared, and the angles at A and B are right angles, so \triangle RAD \sim \triangle RBC. Call the width w; the whole side RB is w + 20. Then \displaystyle \frac{w}{w + 20} = \frac{12}{18}, so 18w = 12w + 240, 6w = 240 and w = 40.
- A proportion has to compare matching sides of \triangle PST and \triangle PQR. ST and QR match, but SQ is not a side of either triangle, so nothing makes \displaystyle \frac{PS}{SQ} equal to the scale ratio. With the numbers of Example 2, \displaystyle \frac{PS}{SQ} = \frac{9}{6} = 1.5, while \displaystyle \frac{ST}{QR} = \frac{6}{10} = 0.6. The correct partner of PS is the whole side PQ = 15, and \displaystyle \frac{9}{15} = 0.6.
- Under a lamp of height H, a person of height p standing d from the post casts a shadow s with \displaystyle \frac{s}{d + s} = \frac{p}{H}, from the similar triangles of Example 4. Clearing fractions, sH = pd + ps, so \displaystyle s = \frac{pd}{H - p}. With p and H fixed, s grows with d: the further person has the longer shadow. Under the sun the rays are parallel, every person’s triangle has the same angles, and shadow over height is the same number for everyone, so equal heights give equal shadows wherever they stand.
- The triangles are similar, \triangle TBC \sim \triangle EDC, with scale factor \displaystyle \frac{DC}{BC}. If DC = BC the scale factor is 1, so every side of \triangle EDC equals its partner in \triangle TBC; in particular DE = TB, the width. (The triangles are then congruent by ASA: the right angles, the sides BC = DC, and the vertical angles at C.)