CHAPTER 3

SIMILARITY

3.1 Similar Triangles

The word similar is used in geometry to describe two figures that have identical shapes but are not necessarily the same size. A working definition of similarity can be obtained in terms of angles and ratios of distances.

Two polygons are similar if corresponding angles are congruent and the ratios of corresponding sides are equal.

Notation. We use the symbol ~ to denote similarity. Thus, we write

to denote that the polygons ABCDE and QRSTU are similar. As with congruency, the order of the letters is important.

To say that ΔABC ~ ΔDEF means that

and

where k is a positive real number.

The constant k is called the proportionality constant or the magnification factor. If k > 1, triangle ABC is larger than triangle DEF; if 0 < k < 1, triangle ABC is smaller than triangle DEF; and if k = 1, the triangles are congruent.

Note that congruent figures are necessarily similar, but similar figures do not need to be congruent.

3.2 Parallel Lines and Similarity

There is a close relationship between parallel lines and similarity.

Lemma 3.2.1. In ΔABC, suppose that D and E are points of AB and AC, respectively, and that DE is parallel to BC, Then

Figures (i), (ii), and (iii) on the following page illustrate the three different possibilities that can occur.

The proof of this lemma uses some simple facts about areas:

Theorem 3.2.2. Given parallel lines l and m and two triangles each with its base on one line and its remaining vertex on the other, the ratio of the areas of the triangles is the ratio of the lengths of their bases.

For the diagram above, the theorem says that

Note that we use square brackets to denote area; that is, [XYZ] is the area of ΔXYZ.

There are two special cases worth mentioning, and these are illustrated by the figures below:

We sketch the proof of Lemma 3.2.1 for case (i). The proof of the other cases is very much the same.

Proof. In case (i), the line DE enters the triangle ABC through side AB and so must exit either through vertex C or through one of the other two sides.¹ Since DE is parallel to BC, the line DE cannot pass through vertex C or any other point on side BC. It follows that DE must exit the triangle through side AC.

Insert segments BE and CD and use the previously cited facts about areas of triangles to get

Since [BDE] = [CDE], we must have

A useful extension of the lemma is the following:

Theorem 3.2.3. Parallel projections preserve ratios. Suppose that l, m, and n are parallel lines that are met by transversals t and t′ at points A, B, C and A′, B′, C′, respectively. Then

Proof. Draw a line through A parallel to t′ meeting m at B″ and n at C″. Then AA′B′B″ and B″B′C′C″ are parallelograms, so AB″ = A′B′ and B″C″ = B′C′. Applying Lemma 3.2.1 to triangle ACC″, we have

and the theorem follows.

Next is the basic similarity theorem for triangles:

Theorem 3.2.4. In ΔABC, suppose that DE is parallel to BC. If D and E are points of AB and AC, with D being neither A nor B nor C, then

Proof. As in the case of Lemma 3.2.1, there are three cases to consider:

We will prove the theorem for case (1). The proofs for the other cases are almost identical. Because of the properties of parallel lines, the corresponding angles of ABC and ADE are equal, so it remains to show that the ratios of the corresponding sides are equal.

By Lemma 3.2.1, we have

which implies that

and this in turn implies that

so that

and therefore

Now. through E, draw EF parallel to AB, with F on BC. Using an argument similar to the one above, we get

Since DE = BF (because BDEF is a parallelogram), we get

Thus,

which completes the proof that triangles ABC and ADE are similar.

There are other conditions that allow us to conclude that two triangles are similar, but there are many occasions where Theorem 3.2.4 is the most appropriate one to use.

Congruency versus Similarity

Congruency and similarity are both equivalence relations — both relations are reflexive, symmetric, and transitive.

Both are reflexive:

ΔABC ≡ ΔABC.

ΔABC ~ ΔABC.

Both are symmetric:

If ΔABC ≡ ΔDEF, then ΔDEF ≡ ΔABC.

If ΔABC ~ ΔDEF, then ΔDEF ~ ΔABC.

Both are transitive:

If ΔABC ≡ ΔDEF and ΔDEF ≡ ΔGHI, then ΔABC ≡ ΔGHI.

If ΔABC ~ ΔDEF and ΔDEF ~ ΔGHI, then ΔABC ~ ΔGHI.

3.3 Other Conditions Implying Similarity

According to the definition, two triangles are similar if and only if the three angles are congruent and the three ratios of the corresponding sides are equal. As with congruent triangles, it is not necessary to check all six items. Here are some of the conditions that will allow us to verify that triangles are similar without checking all six.

Theorem 3.3.1. (AAA or Angle-Angle-Angle Similarity)

Two triangles are similar if and only if all three corresponding angles are congruent.

Exercise 3.3.2. Show that two quadrilaterals need not be similar even if all their corresponding angles are congruent.

Theorem 3.3.3. (sAs or side-Angle-side Similarity)

If in ΔABC and ΔDEF we have

then ΔABC and ΔDEF are similar.

Theorem 3.3.4. (sss or side-side-side Similarity)

Triangles ABC and DEF are similar if and only if

The lowercase letter s in sAs and sss is to remind us that the sides need only be in proportion rather than congruent, while the uppercase letter A is to remind us that the angles must be congruent.

It is worth mentioning that since there are 180° in a triangle, the AAA similarity condition is equivalent to:

Theorem 3.3.5. (AA Similarity)

Two triangles are similar if and only if two of the three corresponding angles are congruent.

Proof of the AAA similarity condition. Make a congruent copy of one triangle so that it shares an angle with the other, that is, so that two sides of the one triangle fall upon two sides of the other. The congruency of the angles then guarantees that the third sides are parallel, and the proof is completed by applying Theorem 3.2.4.

The proof for the sAs similarity condition uses the converse of Lemma 3.2.1.

Lemma 3.3.6. Let P and Q be points on AB and AC with P between A and B and Q between A and C. If

then PQ is parallel to BC.

Proof. Through P, draw a line PR parallel to BC with R on AC. Then, since PR enters triangle ABC through side AB, it must exit the triangle through side AC or side BC. Since PR is parallel to BC, it follows that R is between A and C.

By Lemma 3.2.1,

so it follows that

However, given the positive number k, there is only one point X between A and C such that

and so it follows that R = Q, and so PQ = PR, showing that PQ is parallel to BC.

Proof of the sAs similarity condition.

Suppose that in triangles ABC and DEF we have ∠A ≡ ∠D and that

We may assume that k > 1, that is, that AB > DE and AC > DF. Cut off a segment AP on AB so that AP = DE. Cut off a segment AQ on AC so that AQ = DF.

Then it follows that

and subtracting 1 from both sides of the equation gives us

It now follows from Lemma 3.3.6 that PQ is parallel to BC, and Theorem 3.2.4 implies that ΔABC ~ ΔAPQ. Since ΔAPQ ≡ ΔDEF (by the SAS congruency condition), it follows that ΔABC ~ ΔDEF.

Exercise 3.3.7. Prove that two triangles are similar if they satisfy the sss similarity condition.

3.4 Examples

Pythagoras’ Theorem

Theorem 3.4.1. (Pythagoras’ Theorem)

If two sides of a right triangle have lengths a and b and the hypotenuse has length c, then a² + b² = c².

Proof. In the figure below, drop the perpendicular CD to the hypotenuse, and let AD = p and BD = q.

From the AA similarity condition,

implying that

or, equivalently, that

from which it follows that

and since q + p = c we have a² + b² = c².

Theorem 3.4.2. (Converse of Pythagoras’ Theorem)

Let ABC be a triangle. If

then ∠C is a right angle.

Exercise 3.4.3. Prove the converse of Pythagoras’ Theorem.

Apollonius’ Theorem

As an application of Pythagoras’ Theorem, we prove the following result.

Theorem 3.4.4. (Apollonius’ Theorem)

Let M be the midpoint of the side BC of triangle ABC. Then

Proof. Let AD be the altitude on the base BC, and assume that D lies between M and C.

By Pythagoras’ Theorem,

For other positions of D, the argument is essentially the same.

Example 3.4.5. Use Apollonius’ Theorem to prove that in ΔABC, if m_a = AM is the median from the vertex A, then

where a = BC, b = AC, and c = AB, as in the figure.

Solution. From Apollonius’ Theorem, we have

that is,

so that

Therefore,

Stewart’s Theorem

A related theorem is the following, interesting in its own right, which allows us to express the lengths of the internal angle bisectors of a triangle in terms of the lengths of the sides.

Theorem 3.4.6. (Stewart’s Theorem)

In ΔABC, if D is any point internal to the segment = AD, m = BD, n = DC, b = AC, c = AB, and a = BC, as in the figure, then

Proof. Drop a perpendicular from A to BC, hitting BC at E, and let h = AE and k = DE, as in the figure below.

From Pythagoras’ Theorem, we have

so that

and

Multiplying the equation for c² by n and the equation for b² by m, we have

so that

and since m + n = a, then

Angle Bisector Theorem

Theorem 3.4.7. (The Angle Bisector Theorem)

Let D be a point on side BC of triangle ABC.

Proof. (1) Let l be a line parallel to AD through C meeting AB at E, Then

and

and so ∠AEC ≡ ∠ACE; that is, ΔACE is isosceles. Thus,

and, again using the fact that CE || DA,

Hence,

The proof of (2) using similarity is left as an exercise.

Example 3.4.8. Use Stewart’s Theorem to prove that in ΔABC, if AD = f_A is the internal angle bisector at ∠A, then

where a = BC, b = AC, and c = AB, as in the figure.

Solution. From the Internal Angle Bisector Theorem, we have

so that

and since a = m + n, then

and

That is,

From Stewart’s Theorem, we have

that is,

Therefore,

that is,

Corollary 3.4.9. In ΔABC, if f_A, f_B, and f_C denote the lengths of the internal angle bisectors at ∠A, ∠B, and ∠C, respectively, then

Exercise 3.4.10. In ΔABC, let AD, BE, and CF be the angle bisectors of ∠BAC, ∠ABC, and ∠ACB, respectively. Show that if

then

Medians

In Chapter 2, we used area to show that the medians of a triangle trisect each other and are concurrent. Here is how similar triangles can be used to prove the same thing:

Example 3.4.11. Let BE and CF be two medians of a triangle meeting each other at G. Show that

and deduce that all three medians are concurrent at G.

Solution. In the figure below,

by the sAs similarity condition we have ΔABC ~ ΔAFE, with proportionality constant 1/2. This means that ∠ABC ≡ ∠AFE, from which it follows that FE is parallel to BC with FE = BC/2. This means that ΔBCG ~ ΔEFG with a proportionality constant of 1/2. Hence,

which means that EG/GB = FG/GC = 1/2.

Applying the same reasoning, medians BE and AD intersect at a point H for which EH/HB = HD/AH = 1/2. But this means that

and since both G and H are between B and E, we must have H = G. This shows that the three medians are concurrent at G.

We will end this section with a theorem that is sometimes useful in construction problems.

Theorem 3.4.12. In ΔABC, suppose that D and E are points on AB and AC, respectively, such that DE || BC. Let P be a point on BC between B and C, and let Q be a point on DE between D and E. Then A, P, and Q are collinear if and only if

The figures below illustrate the three cases that can arise.

Proof.

(i) Suppose A, P, and Q are collinear. Since DE is parallel to BC, we have

and so

from which it follows that

or, equivalently, that

(ii) Conversely, suppose that

Draw the line AP meeting DE at Q′. We will show that Q = Q′.

Since A, P, and Q′ are collinear, the first part of the proof shows that

We are given that

so that

from which it follows that Q′ = Q, and this completes the proof.

3.5 Construction Problems

If you were asked to bisect a given line segment, you would probably construct the right bisector. How would you solve the following problem?

Example 3.5.1. To divide a given line segment into three equal parts.

Many construction problems involve similarity. The idea is to create a figure similar to the desired one and then, by means of parallel lines or some other device, transform the similar figure into the desired figure.

Creating a similar figure effectively removes size restrictions, and this can make a difficult problem seem almost trivial. Without size restrictions, Example 3.5.1 becomes:

To construct any line segment that is divided into three equal parts.

This is an easy task: using any line, fix the compass at any radius, and strike off three segments of equal length AB, BC, and CD, Then AD is a line segment that has been divided into three equal parts. What follows shows how this can be used to solve the original problem.

Solution. We are given the line segment PQ, which is to be divided into three equal parts. First, draw a ray from P making an angle with PQ. With the compass set at a convenient radius, strike off congruent segments PR′, R′S′, and S′Q′ along the ray.

Join Q′ and Q. Through R′ and S′, draw lines parallel to Q′Q that meet PQ at R and S. Since parallel projections preserve ratios, we have

and so RS = SQ. Similarly, PR = RS.

The preceding construction can be modified to divide a line segment into given proportions. That is, it can be used to solve the following: given AB and line segments of length p and q, construct the point C between A and B so that

There is frequently more than one way to use similarity to solve a construction problem.

Example 3.5.2. Given two parallel lines l and m and given points A on l, B on m, and P between l and m, construct a line n through P that meets l at C and m at D so that

as in figure (a) below.

Solution.

Example 3.5.3. Given a point P inside an angle A, construct the two circles passing through P that are tangent to the arms of the angle.

Solution. In the diagram, several construction arcs have been omitted for clarity.

Analysis Figure:

Construction:

Justification:

To show that C(O, OQ) is one of the desired circles, we know that it is tangent to both arms of the angle because O is on the angle bisector and the radius OQ is perpendicular to an arm of the angle. It remains to show that the circle passes through P; that is, that OP = OQ, the radius of the circle.

Since O′Q′ || OQ, it follows that

Since Q′P′ || QP, it follows that

and, consequently, that

Again, since Q′P′ || QP, it follows that ∠AQ′P′ = ∠AQP, and so

Then, by the sAs criteria, ΔO′Q′P′ ~ ΔOQP, and so

Thus OP = OQ, and this completes the proof.

Remark. Example 3.5.3 uses the idea that given a point A and a figure F, we can construct a similar figure G by constructing all points Q such that AQ = k · AP for points P in F, where k is a fixed nonzero constant. In the example, we constructed a circle centered at a point O′ and then constructed a magnified version of this circle centered at point O.

The transformation that maps each point P to the corresponding point Q so that AQ = k · AP is called a homothety and is denoted H(A, k). The number k is the magnification constant. The figure on the following page illustrates the effect of H(A,3).

Here are two problems that are solved by translations rather than a homothety. The strategy is very similar: we construct a circle that fulfills part of the criteria and then use a translation to move it to the correct place.

Example 3.5.4. Given a line l and points P and Q such that P and Q are on the same side of l and such that PQ ⊥ l, construct the circle tangent to l that passes through P and Q.

Solution. In the figure, we have omitted construction lines for the standard constructions (like dropping a perpendicular from a point to a line).

Analysis Figure:

Construction:

Justification:

PP′O′O is a parallelogram because PP′ is congruent to and parallel to OO′. Then OP = O′P′ = r, showing that P is on the circle C(O, r). Since O is on the right bisector of PQ. it follows that OQ = OP = r, showing that Q is on C(O, r).

Finally, since m and l are parallel, the perpendicular distance from O to l is the same as the perpendicular distance from O′ to l, and it follows that C(O, r) is tangent to l.

Example 3.5.5. Given two disjoint circles of radii R and r, construct the four tangents to the circle.

The diagram on the right illustrates that two nonoverlapping circles have four common tangent lines. The problem is to construct those lines. We will show how to construct the “external” tangents and leave the construction of the “internal” ones as an exercise.

Solution.

Analysis Figure:

Let us suppose that the radii of the circles are R and r, as shown on the right. If we draw a line m through the center Q of the smaller circle parallel to a tangent line l as shown, then its distance from the center O of the larger circle will be R − r, and it will be tangent to C(O, R − r).

Construction:

Here is the step-by-step construction.

Justification:

Let QU be parallel to SP so that QSPU is a rectangle. Then QU = PS = r where r is the radius of the smaller circle. Since PU ⊥ QU, then PU is tangent to the smaller circle.

Since OP is a radius of the larger circle and since PU ⊥ OP, then PU is tangent to the larger circle.

Let QV be parallel to TW so that QTWV is a rectangle. Reasoning as before, WV is tangent to both circles.

3.6 The Power of a Point

The following problem is taken from AHA! Insight, a delightful book written by Martin Gardner. Gardner attributes the problem to Henry Wadsworth Longfellow.

A lily pad floats on the surface of a pond as far as possible from where its root is attached to the bottom. If it is pulled out of the water vertically, until its stem is taut, it can be lifted 10 cm out of the water. The stem enters the water at a point 50 cm from where the lily pad was originally floating. What is the depth of the pond?

Most people attack this problem by using Pythagoras’ Theorem together with the fact that a chord is perpendicular to the radius that bisects it. At the end of this section we will see that there is a much more elegant approach. We begin with a valuable fact about intersecting secants and chords.

Theorem 3.6.1. Let ω be a circle, let P be any point in the plane, and let l and m be two lines through P meeting the circle at A and B and C and D, respectively. Then

Proof. If P is outside the circle, suppose that A is between P and B and that C is between P and D. Insert the line segments AD and BC, and consider triangles PAD and PCB.

By Thales’ Theorem,

so by the AA similarity criteria, ΔPAD ~ ΔPCB.

Consequently,

and so

If P is inside the circle, again insert line segments AD and BC. Then, triangles PDA and PBC are similar, and so we again have PA · PB = PC · PD.

When P is outside the circle, an interesting thing happens if we swing the secant line PCD to a tangent position, as in the figure on the following page. In this case, the points C and D approach each other and coalesce at the point T, and so the product PC · PD approaches PT².

As a consequence:

Theorem 3.6.2. Let P be a point in the plane outside a circle ω. Let PT be tangent to the circle at T, and let l be a line through P meeting the circle at A and B. Then

Proof. Here is a proof that does not involve limits.

We may assume that A is between P and B. Insert the segments TA and TB, and we have the figure below.

A consequence of Thales’ Theorem is that ΔPTA and ∠PBT are equal in size. It follows that ΔPTA ~ ΔPBT, and so

It follows immediately that PT² = PA · PB.

If P is inside the circle, then, of course, no tangent to the circle passes through P. However, there is a result that looks somewhat like the previous theorem when P is inside the circle:

Theorem 3.6.3. Let P be a point in the plane inside a circle ω. Let TS be a chord whose midpoint is P and let AB be any other chord containing P. Then

Now let l be a line, and assign a direction to the line. For two points A and B on the line, with A ≠ B, let AB be the distance between A and B. The directed distance or signed distance from A to B, denoted , is defined as follows:

By using directed distances, the theorems above can be combined and extended as follows:

Theorem 3.6.4. Let P be any point in the plane, let ω be a given circle, and let l be a line through P meeting the circle at A and B. Then the value of the product is independent of the line l, and

The value is called the power of the point P with respect to the circle w.

Suppose we are given a circle ω with center O and radius r, and suppose that we are given a point P and that we know the distance OP. It is fairly obvious how we could experimentally determine the power of P with respect to ω draw any line through P meeting the circle at A and B, and measure PA and PB. Of course, if one chooses a line passing through O, then PA and PB are readily determined without recourse to measurement (see the figure on the following page).

By doing this, we find:

Corollary 3.6.5. The power of a point P with respect to a circle with center O and radius r is OP² − r².

The Lily Pad Problem

The figure below illustrates the geometry of the lily pad problem. The depth of the pond is OP = x, and the length of the stem of the lily pad is the radius of a circle centered at O.

We know now that PA · PB = PT², so referring to the diagram,

and therefore the pond is 120 cm deep.

Example 3.6.6. A person whose eyes are exactly 1 km above sea level looks towards the horizon at sea. The point on the horizon is 113 km from the observer’s eyes. Use this to estimate the diameter of the earth.

Solution. The line of sight to the horizon is tangent to the earth. If x is the diameter of the earth, we have

so

3.7 Solutions to the Exercises

Solution to Exercise 3.3.2

There are many examples, as the following figures indicate:

Solution to Exercise 3.3.7

We are given that

and so we have to prove that the angles are equal.

We may assume that k < 1, that is, that AB < DE.

Let P be the point on ED such that EP = BA. Draw PQ parallel to DF with Q on EF. By Theorem 3.2.4, ΔEPQ ~ ΔEDF, and so

This implies that EQ = BC.

Similarly, PQ = AC, so ΔEPQ ≡ ΔBAC, and it follows that

Since PQ || DF we also have

and this completes the proof.

Solution to Exercise 3.4.3

Given triangle ABC with AB² = BC² + CA², let PQR be a triangle with a right angle at R, such that QR = BC and RP = CA. Then

Hence, PQ = AB, so ABC and PQR are congruent triangles by SSS. It follows that ∠C is a right angle.

Solution to Exercise 3.4.10

Let β = ∠CBE and let γ = ∠BCF. First, we note that if α = β, then ΔABC is isosceles, so that BE = CF. Thus, we may assume, without loss of generality, that β < γ.

Let G be the point on BE between B and E such that ∠FCG = β. Then

so by Thales’ Theorem, F, G, B, and C lie on a circle.

Now,

so that ∠CBF < ∠BCG < 90, and therefore CF < BG < BE.

These last inequalities follow from the fact that given two chords in a circle, one is smaller if and only if it is further from the center, and this is true if and only if it subtends a smaller central angle (the SAS inequality and its converse) and therefore a smaller acute angle at the circumference.

So, if a triangle has two different angles, the smaller angle has the longer internal bisector, which is what we wanted to show.

3.8 Problems

¹ This statement is known as Pasch’s Axiom.