Geometry · Chapter 27

Similar Triangles

Same shape, different size: spot hidden similar triangles, use their side ratios, and square the ratio for areas.

What's this chapter about?

27.1 Congruent Triangles
27.2 Similar Triangles
27.3 Practice Problems

You'll need

27.10%

Congruent Triangles

In this section you will be able to decide when two triangles must be exact copies, and use a congruent copy to move a length where you need it.

You cut a paper triangle and your friend wants an exact copy. You may only tell them three facts. Which three facts are enough, and which three can fool them?

Two triangles are congruent when one is an exact copy of the other. It may be turned or flipped, but every side and every angle matches. Then everything else matches too: area, perimeter, all of it.

You don’t need to check all six parts. Three well-chosen facts are enough. But watch out for SSA: it can fit two different triangles.

Look:

A congruent copy lets you move a length to a better spot. In Example 27.1 you turn one triangle a quarter turn around B, and two short segments line up into one long one.

Exploreplay first, formulas later

Try this: Press SSS, then SAS, then ASA. How many triangles fit the given data each time?
Try this: Press SSA. Why are there two triangles now? Look at where C1 and C2 sit.
Try this: In SSA, change angle A with the stepper. Can you find an angle where only one triangle fits, or none?
Try this: Press AAA. The angles match, but the sizes don't. Is that enough for congruent?

Name it

Theorem 27.1.1

For congruent triangles:

All the angles of the triangles are same
All corresponding sides are equal
Area, perimeter, inradius, circumradius, etc. are equal

Concept 27.1.2 · Congruence Test

Two triangles are congruent if the three angles and sides in the triangle are the same. In other words, the triangles are the same (not necessarily same orientation). In general, triangles are congruent if:

AAS congruence: Two angles of the triangles are same and the 1 side next to the 2 angles are equal
SAS Congruence (Side Angle Side): Two sides are equal and the angle between the sides are equal
SSS congruence (Side Side Side): All three sides are congruent
HL congruence (Hypotenuse Leg): In a right triangle, hypotenuse and leg are equal
LL congruence (LL Leg): In a right triangle, the two legs are equal

Warning: SSA is not a valid triangle congruence: if 2 corresponding sides are congruent and the angle which is not between the 2 congruent sides is equal, then the triangles are not necessarily congruent

Watch itpredict, then reveal

Example 27.1

EMCC

In square $ABCD$ , points $E$ and $F$ lie on segments $AD$ and $CD$ , respectively. Given that $\angle EBF = 45^\circ$ , $DE = 12$ , and $DF = 35$ , compute $AB$ .

1
Triangle $EDF$ has a right angle at $D$ . How long is $EF$ ?

Predict first:
Your prediction

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Contest cornerreal contest problems

Problem 27.3.4

AMC 8

One-inch squares are cut from the corners of this 5 inch square. What is the area in square inches of the largest square that can fit into the remaining space?

Problem 27.3.5

AMC 10

A square piece of paper has side length $1$ and vertices $A, B, C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$ , and edge $\overline{AB}$ at point $E$ . Suppose that $C'D = \frac{1}{3}$ . What is the perimeter of triangle $\triangle AEC'$ ?

Problem 27.3.9(additional)

EMCC

Let $ABCD$ be a convex quadrilateral with $AB = BC = CA$ . Suppose that point $P$ lies inside the quadrilateral with $AP = PD = DA$ and $\angle PCD = 30^\circ$ . Given that $CP = 2$ and $CD = 3$ , compute $CA$ .

Checkpoint for section 27.1

Get 2 of 3 right to master this section. One try each, no hints.

27.20%

Similar Triangles

In this section you will be able to spot similar triangles, match up their sides with one scale factor, and square that factor for areas.

A lamp post casts a long shadow, and so do you. You never climb the post, yet you can find its height with a tape measure. How? Your body and the post make two triangles with the same shape.

Two triangles are similar when they have the same angles. One is a scaled copy of the other: every side is multiplied by the same scale factor k.

The easiest place to find them: a line parallel to one side of a triangle cuts off a smaller, similar triangle. Two parallel lines crossed by an X make an hourglass of similar triangles.

Look:

Lengths scale by k, but areas scale by k squared. Double every side and the area becomes 4 times as big.

Exploreplay first, formulas later

Try this: Drag the person (or use the stepper). The person's height divided by the post's height: is it the same as TF divided by TP?
Try this: Move the person so TF : TP = 1/2. How tall is the person compared with the post?
Try this: Watch the area ratio. When the side ratio is 1/2, why is the area ratio 1/4, not 1/2?

Name it

Concept 27.2.1 · Similarity Test

Two triangles are similar if the three angles in the triangle are the same. In other words, the triangles are the same shape multiplied by a scale factor.

In general, triangles are similar if:

AA similarity: Two angles of the triangles are same, which basically means that the third angle will be equal)
SAS similarity (Side Angle Side): Two sides are proportional and the angle between the sides is equal
SSS similarity (Side Side Side): All three sides are proportional
HL similarity (Hypotenuse Leg): In a right triangle, the hypotenuse and leg are proportional
LL similarity (LL Leg): In a right triangle, the two legs are proportional

Warning: SSA does not mean triangles are similar

An easy way to detect similar triangles is if bases of triangles are parallel and the sides of the triangles are collinear (see the hourglass figure below).

The hourglass: the two triangles that meet at the crossing point are similar.

Theorem 27.2.2

For similar triangles:

All the angles of the triangles are same
All corresponding sides have same ratio
Area ratio is the square of side length ratio

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Example 27.2

EMCC

In triangle $ABC$ , $\angle BAC = 90^\circ$ , and the length of segment $AB$ is 2011. Let $M$ be the midpoint of $BC$ and $D$ the midpoint of $AM$ . Let $E$ be the point on segment $AB$ such that $EM \parallel CD$ . What is the length of segment $BE$ ?

1
Let line $CD$ meet $AB$ at $F$ . In triangle $AME$ , segment $DF$ is parallel to $ME$ . What is $AD : AM$ ?

Predict first:
Your prediction

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Example 27.3

EMCC

In triangle $ABC$ with $BC = 5$ , $CA = 13$ , and $AB = 12$ , Points $E$ and $F$ are chosen on sides $AC$ and $AB$ , respectively, such that $EF \parallel BC$ . Given that triangle $AEF$ and trapezoid $EFBC$ have the same perimeter, find the length of $EF$ .

1
Why is triangle $AFE$ similar to triangle $ABC$ ?

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Example 27.4

EMCC

In $\triangle ABC$ , $AC = 8$ and $AC < AB$ . Point $D$ lies on side $BC$ with $\angle BAD = \angle CAD$ . Let $M$ be the midpoint of $BC$ . The line passing through $M$ parallel to $AD$ intersects lines $AB$ and $AC$ at $F$ and $E$ , respectively. If $EF = \sqrt{2}$ and $AF = 1$ , what is the length of segment $BC$ ?

1
$FE \parallel AD$ , and $AD$ splits angle $A$ into two equal angles. What does that say about $\angle AFE$ and $\angle AEF$ ?

Predict first:
Your prediction

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Try it

Problem 27.3.1

AMC 8

The diagram shows an octagon consisting of 10 unit squares. The portion below $\overline{PQ}$ is a unit square and a triangle with base 5. If $\overline{PQ}$ bisects the area of the octagon, what is the ratio $\dfrac{XQ}{QY}$ ?

Problem 27.3.3

AMC 8

Point $E$ is the midpoint of side $\overline{CD}$ in square $ABCD$ , and $\overline{BE}$ meets diagonal $\overline{AC}$ at $F$ . The area of quadrilateral $AFED$ is $45$ . What is the area of $ABCD$ ?

Contest cornerreal contest problems

Problem 27.3.2

AMC 8

Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC = CB = BA = 1$ . The area of the "bat wings" (shaded area) is

Problem 27.3.6

AMC 10B

Right $\triangle ABC$ has $AB = 3$ , $BC = 4$ , and $AC = 5$ . Square $XYZW$ is inscribed in $\triangle ABC$ with $X$ and $Y$ on $\overline{AC}$ , $W$ on $\overline{AB}$ , and $Z$ on $\overline{BC}$ . What is the side length of the square?

Problem 27.3.8(additional)

EMCC

In trapezoid $ABCD$ , points $E$ and $F$ lie on sides $BC$ and $AD$ , respectively, such that $AB \parallel CD \parallel EF$ . Given that $AB = 3$ , $EF = 5$ , and $CD = 6$ , the ratio $[ABEF] / [CDEF]$ can be written as $\frac{a}{b}$ , where $a$ and $b$ are relatively prime positive integers. Find $a + b$ . (Note: $[F]$ denotes the area of $F$ .)

Problem 27.3.10(additional)

BmMT

Let $ABC$ be a right triangle with hypotenuse $AB$ such that $AC = 36$ and $BC = 15$ . A semicircle is inscribed in $ABC$ as shown, such that the diameter $XC$ of the semicircle lies on side $AC$ and that the semicircle is tangent to $AB$ . What is the radius of the semicircle?

Problem 27.3.11(additional)

AMC 10/12

A square with side length $x$ is inscribed in a right triangle with sides of length $3$ , $4$ , and $5$ so that one vertex of the square coincides with the right-angle vertex of the triangle. A square with side length $y$ is inscribed in another right triangle with sides of length $3$ , $4$ , and $5$ so that one side of the square lies on the hypotenuse of the triangle. What is $\dfrac{x}{y}$ ?

Checkpoint for section 27.2

Get 2 of 3 right to master this section. One try each, no hints.

A triangle has area $5$ . Every side is tripled. What is the area of the new triangle?

Two similar triangles have areas $6$ and $54$ . The shortest side of the small one is $3$ . How long is the shortest side of the big one?

27.3

Practice Problems

In this section you will be able to mix congruent and similar triangles on contest problems.

Time to hunt for hidden triangles! In every problem, look for parallel lines, right angles and shared angles. Every problem has three hints if you get stuck.

Want every problem from this chapter on one page? Open the problem set.

Contest cornerreal contest problems

Problem 27.3.7(additional)

BmMT

Let $ABCDEFGHIJKL$ be the equilateral dodecagon shown below, and each angle is either $90^\circ$ or $270^\circ$ . Let $M$ be the midpoint of $CD$ , and suppose $HM$ splits the dodecagon into two regions. The ratio of the area of the larger region to the area of the smaller region can be expressed as $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Compute $m + n$ .