Similar Triangles
1. What This Chapter Covers
Snigdha wants to know the height of a tall tree in her backyard. Her uncle asks her to fetch a mirror, places it flat on the ground some distance from the tree, and tells her to stand where she can just see the treetop in it.
Drawing the girl, the mirror and the tree gives two triangles, ABC and DEC. They are not congruent — the tree is far bigger than Snigdha — but they have the same shape. Figures of the same shape that need not be the same size are called similar figures.
That single idea is how the heights of trees, towers and mountains, and the distances of the Sun and Moon, have all been found. None of them can be reached with a measuring tape; all of them yield to indirect measurement based on similarity.
The index allots this chapter 18 periods across July and August — the largest allocation in the book — and it runs from textbook page 195 to page 228, with four numbered exercises and an Optional Exercise.
2. Same Shape Is Not the Same as Alike
Section 8.2 makes the definition precise by showing what fails it. Take a drawing of a car. Keep its breadth and double its length, and you get a second picture; keep its length and double its breadth, and you get a third.
A photographer makes the same point the other way. Printing a 35 mm negative at 45 mm enlarges every line segment in the ratio 35 : 45, and the angles are unchanged, so the two photographs are similar.
So the definition has two halves, and the book insists that one alone is not sufficient. Two polygons with the same number of sides are similar if
- all their corresponding angles are equal, and
- all their corresponding sides are in the same ratio.
A square and a rectangle satisfy the first and fail the second, so they are not similar. A square and a rhombus satisfy the second and fail the first.
The common ratio is the scale factor, and the book records what it means: with K > 1 the figure is enlarged, with K = 1 it is congruent, and with K < 1 it is reduced. Every congruent pair is therefore similar, but similar figures need not be congruent.
All regular polygons with the same number of sides are similar, and so are all circles — they differ only in size.
3. The Basic Proportionality Theorem
For triangles, the two conditions collapse into one. That is the surprise of this chapter, and it rests on a result the book reaches through a ruled-paper activity: draw a triangle with its base on one line, and any other line of the page cuts the two remaining sides in the same ratio.
Theorem 8.1, the Basic Proportionality Theorem, states it: if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. It is also called Thales Theorem.
The proof is a neat piece of reasoning that uses area rather than length. Joining BE and CD and dropping perpendiculars DM and EN gives ar(ADE)/ar(BDE) = AD/DB, because the two triangles share the same height from E. The same argument on the other side gives ar(ADE)/ar(CDE) = AE/EC.
The last step is the one to remember. Triangles BDE and CDE sit on the same base DE and between the same parallels, so their areas are equal. The two ratios therefore have equal denominators, and AD/DB = AE/EC follows.
Theorem 8.2 is the converse: if a line divides two sides of a triangle in the same ratio, that line is parallel to the third side. Its proof is by contradiction — assume a different line through D is the parallel one, and show that its intersection point must coincide with E.
The converse is what turns the theorem into a tool. Example-3 uses it to prove that a quadrilateral whose diagonals cut each other in equal ratios must be a trapezium, and Example-4 uses it on a trapezium to show that a line parallel to the parallel sides divides both slanted sides alike.
Section 8.3 also uses Thales for a construction: dividing a segment in a given ratio without measuring it. Draw a ray at an acute angle, step off m + n equal lengths along it with a compass, join the last point to the far end of the segment, and draw a parallel through the mth point.
Example-1 is the plainest use. In a triangle with DE ∥ BC, AD/DB = 3/5 and AC = 5.6 cm; since AE/EC must also be 3/5 and AE + EC = 5.6, solving AE/(5.6 − AE) = 3/5 gives AE = 2.1 cm.
Example-2 is the same theorem with algebra attached. With LM ∥ AB, AL = x − 3, AC = 2x, BM = x − 2 and BC = 2x + 3, the theorem gives (x − 3)/(x + 3) = (x − 2)/(x + 5). Cross-multiplying cancels the x² terms and leaves x = 9. Notice what had to be computed first: LC is AC − AL, not AC, and MC is BC − BM.
4. The Three Criteria
For polygons in general you must check angles and sides. For triangles you need only one of the two, and the chapter proves it three times over.
Theorem 8.3 (AAA) proves that equal corresponding angles force proportional sides. The construction is worth following: mark P on DE and Q on DF so that DP = AB and DQ = AC, join PQ, and show that triangle ABC is congruent to triangle DPQ. That makes PQ parallel to EF, and Thales finishes the job.
Since the angles of a triangle add to 180°, two equal pairs force the third, so AAA reduces to AA — and the book states the criterion in that shorter form.
Theorem 8.4 (SSS) goes the other way: proportional sides force equal angles. Theorem 8.5 (SAS) needs only one angle, provided it is the angle between the two proportional sides.
The book draws the moral explicitly. For polygons in general, one condition is not enough; for triangles, one automatically implies the other.
Section 8.4 closes with a second construction: building a triangle similar to a given one at a stated scale factor, by stepping off equal lengths along a ray and drawing parallels.
5. What Similar Triangles Are For
Example-5 is the classic. A person 1.65 m tall casts a shadow 1.8 m long; a lamp post nearby casts a shadow of 5.4 m. How tall is the post?
Both triangles have a right angle at the ground, and because the sun's rays are parallel at any instant, the angles of elevation are equal too. AA similarity gives 1.65/PQ = 1.8/5.4, so the post is 4.95 m tall.
Example-6 turns the chapter's opening mirror trick into arithmetic: a man 1.5 m tall standing 0.4 m from a mirror that is 87.6 m from a tower gives a tower height of 328.5 m. Example-7 is a privacy problem — how high a fence must be raised to block a neighbour's view — and the answer is 1.8 m.
What makes these work is always the same pair of facts: a right angle shared by both triangles, and a second angle equal for a physical reason. For shadows it is that sunlight arrives in parallel rays; for the mirror it is that the angle of incidence equals the angle of reflection, so their complements are equal too.
Exercise 8.2 turns the idea loose on a moving object. A girl 90 cm tall walks away from a 3.6 m lamp post at 1.2 m per second; after 4 seconds she is 4.8 m from its base, and similar triangles give her shadow as 1.6 m. Her shadow grows steadily as she walks, because the ratio of heights, 90 : 360, never changes.
6. Areas Grow as the Square
If the sides of one triangle are twice those of another, what happens to the area?
Theorem 8.6 answers it: the ratio of the areas of two similar triangles equals the ratio of the squares of their corresponding sides. The proof writes each area as half base times altitude, then uses AA similarity on the two right triangles formed by the altitudes to show that the altitudes are in the same ratio as the sides. The ratio of sides therefore appears twice, once from the base and once from the height.
Two consequences are worth naming. Example-8 shows that similar triangles of equal area must be congruent, since the ratio of areas being 1 forces the ratio of sides to be 1. And the midpoint triangle of any triangle has exactly one quarter of its area, because its sides are half as long.
Exercise 8.3 question 2 is a good test of the idea in reverse: if XY ∥ AC divides triangle ABC into two parts of equal area, then the ratio of sides is 1 : √2, and AX/XB works out to √2 − 1.
7. Pythagoras, Proved by Similarity
The chapter's last main result is one students already know, proved in a way they probably do not.
Theorem 8.7 says that dropping a perpendicular from the right angle to the hypotenuse produces two triangles each similar to the original, and therefore to each other. Both share an angle with the whole triangle and both have a right angle, so AA does the work.
Theorem 8.8 then follows in four lines. From the first similarity, AD × AC = AB²; from the second, CD × AC = BC². Adding them gives AC(AD + CD) = AB² + BC², and since AD + CD is just AC, the result is AC² = AB² + BC².
The book records that this was given earlier by the Indian mathematician Baudhayana, about 800 BCE, in the form "the diagonal of a rectangle produces by itself the same area as produced by its both sides", and calls it the Baudhayana theorem alongside its Greek name.
Theorem 8.9 proves the converse by constructing a right triangle with two matching sides and showing the original must be congruent to it.
A Think and Discuss box sets a genuinely interesting puzzle: for a right triangle with integer sides, at least one of the measurements must be even. The reason is a parity argument — if both legs were odd, the sum of their squares would leave remainder 2 on division by 4, and no square does that.
Exercise 8.4 is the longest in the chapter and its questions repay sorting into families. Three are pure identities: the sum of the squares of a rhombus's sides equals the sum of the squares of its diagonals; three times the square of an equilateral triangle's side equals four times the square of its altitude; and an isosceles right triangle has AB² = 2AC².
Four more use the altitude-to-hypotenuse result in disguise, asking for PM² = QM · MR or AB² = BC · BD. Three are word problems: the wire and the pole, the two poles, and two aeroplanes leaving an airport at right angles.
Example-14 shows the chapter meeting chapter 5. A right triangle whose hypotenuse is 6 m more than twice the shortest side, and whose third side is 2 m less than the hypotenuse, gives the quadratic x² − 8x − 20 = 0, so the sides are 10 m, 24 m and 26 m.
8. What the Exercises Ask, and What the Book Answers
Exercise 8.1 has nine questions on the Basic Proportionality Theorem, all but one of them proofs. Exercise 8.2 has thirteen mixing similarity criteria with applications and three constructions. Exercise 8.3 has six on areas. Exercise 8.4 has fourteen on Pythagoras, and the Optional Exercise adds six.
The printed answers sit at textbook pages 382 and 383, and there are few of them, because most questions ask for proofs or constructions. Every printed answer for this chapter is correct — the only chapter in the book so far of which that can be said without qualification.
The ones worth checking your own work against, from Exercise 8.2: the trapezium question gives x = 5 cm and y = 2.8125 cm; the girl walking away from a lamp post has a shadow of 1.6 m after 4 seconds; the flag pole comparison gives a building 16 m tall; and two triangles of perimeters 30 and 20 give a corresponding side of 8 cm.
From Exercise 8.3: the midpoint triangle has area ratio 1 : 4, a pair of similar triangles with BC = 3 and EF = 4 scales an area of 54 up to 96, and areas of 81 and 49 with the larger altitude 4.5 give a smaller altitude of 3.5.
From Exercise 8.4: a 24 m wire on an 18 m pole reaches a stake 6√7 m away, poles of 6 m and 11 m with feet 12 m apart have tops 13 m apart, and the isosceles right triangle question gives an area ratio of 1 : 2.
9. Two Things in the Chapter Text
The first sits inside the definition the chapter exists to teach. Page 196, having just stretched a car two different ways and called both figures distorted, asks whether they are similar and answers: "No, they have same shape, yet they are not similar."
That contradicts the chapter's own definition twice over. Similar figures are precisely those with the same shape, so nothing can have the same shape and fail to be similar. And the stretched cars do not have the same shape — that is exactly what "distorted" means. The sentence should say that they do not have the same shape, and are therefore not similar.
The second is in section 8.7, on the forms of theoretical statements. Page 226 offers this worked example of negation:
p : All irrational numbers are real numbers. ~p : All irrational numbers are not real numbers.
That is not the negation. The negation of "all A are B" is "not all A are B", or equivalently "some A is not B". What the page gives is the much stronger claim that no irrational number is real — the contrary of p, not its contradictory.
The two happen to agree in truth value here, because p is true and both proposed negations are false, so the error is invisible in this example. It stops being invisible the moment a statement is only partly true: negating "all prime numbers are odd" as "all prime numbers are not odd" turns one false statement into another, when the correct negation is true.
Two smaller slips: page 197 heads a box "Think and Dissuss", and page 198 offers "Similar fgures".
One item of wording is worth a glance too. The Do This box on page 198 asks students to fill the blank in "Two polygons with same number of sides are ....... if their corresponding angles are equal and corresponding sides are equal", where the definition on the facing page requires the sides to be in the same ratio. Equal sides together with equal angles would make the polygons congruent, not merely similar.
10. Summary
Similar figures have the same shape but need not have the same size. For polygons this needs two conditions together — equal corresponding angles and corresponding sides in the same ratio — and either one alone is not enough.
The Basic Proportionality Theorem says a line parallel to one side of a triangle divides the other two sides in the same ratio, and its converse says that a line dividing two sides in the same ratio must be parallel to the third. Both are proved through areas of triangles on the same base between the same parallels.
For triangles the two similarity conditions become one. AA is enough, because the third angle follows from the angle sum; SSS is enough, because proportional sides force equal angles; and SAS is enough provided the equal angle lies between the two proportional sides.
The ratio of the areas of two similar triangles is the square of the ratio of their sides, because the ratio enters once through the base and once through the height. Similar triangles of equal area are therefore congruent.
Dropping a perpendicular from the right angle of a right triangle to the hypotenuse makes three similar triangles, and Pythagoras theorem falls out of two of the resulting proportions. The book names it the Baudhayana theorem as well, after the Indian statement of it from about 800 BCE.
