Proportionality Theorems for Triangles

In this lesson, the similarity of triangles will be used to prove some claims about triangles.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

The Pythagorean Theorem
The concept of similarity.
Conditions for similarity of polygons.
Conditions that guarantee the similarity of triangles.

Try your knowledge on these topics.

Which of the following conditions guarantee that two triangles are similar?

Challenge

Solving Problems Using Triangle Similarity

The street lamps are $15$ and $10$ feet tall. How tall is the Grim Reaper?

Explore

Investigating Triangle Proportionality

Move the point on the side of the triangle. The applet draws a line parallel to another side of the triangle and gives the length of four line segments.

Move the vertices and investigate the relationship between these lengths. What do you notice?
Explore different triangles in relation to the pyramids!

Discussion

Triangle Proportionality Theorem

The previous exploration can lead to discovering the following claim, which is often referred to as the Side-Splitter Theorem.

Rule

Triangle Proportionality Theorem

If a segment parallel to one of the sides of a triangle is drawn between the other sides, the segment divides the other two sides proportionally.

Triangle ABC with segment DE drawn parallel to AB

Based on the diagram, the following relation holds true.

If $\overline{D E} ∥ \overline{A B},$ then $\frac{A D}{D C} = \frac{B E}{E C}$

Proof

Since $\overline{D E}$ and $\overline{A B}$ are parallel, by the Corresponding Angles Theorem, $∠ C D E$ and $∠ C A B$ are congruent. Similarly, $∠ C E D$ and $∠ C B A$ are congruent.

Therefore, by the Angle-Angle Similarity Theorem,

△ A B C

and

△ D E C

are similar. Consequently, their corresponding sides are proportional.

△ A B C \sim △ D E C ⇓ \frac{A C}{D C} = \frac{B C}{E C}

Applying the Segment Addition Postulate, both numerators can be rewritten.

A C B C = A D + D C = B E + E C

Substituting these expressions into the equation above, the required proportion will be obtained.

\frac{A C}{D C} = \frac{B C}{E C}

SubstituteII

$A C = A D + D C$ , $B C = B E + E C$

\frac{A D + D C}{D C} = \frac{B E + E C}{E C}

WriteSumFrac

Write as a sum of fractions

\frac{A D}{D C} + \frac{D C}{D C} = \frac{B E}{E C} + \frac{E C}{E C}

SimpQuot

Simplify quotient

\frac{A D}{D C} + 1 = \frac{B E}{E C} + 1

SubEqn

$LHS - 1 = RHS - 1$

\frac{A D}{D C} = \frac{B E}{E C}

Pop Quiz

Practice Using the Triangle Proportionality Theorem

Example

Using the Triangle Proportionality Theorem to Draw Segments

In the following example the Triangle Proportionality Theorem can be used after rearranging the segments to form triangles. Given the segments on the diagram, construct a segment of length $a b .$

Vertical Stacked Segments: Lengths of b, a, and 1 unit from top to bottom.

Answer

Hint

Use that $a b : b = a : 1 .$

Solution

Move the slider to rearrange the given line segments. Note that this can also be done on paper using a straightedge to draw straight lines, followed by using a compass to copy the segments.

Label the points on this rearranged graph, connect the endpoints of the segments of length $1$ and $b,$ and draw a parallel line to this connecting line through the endpoint of the segment of length $a .$

△ A C E

the transversal

B D

is parallel to the side

\overline{C E} .

According to the Triangle Proportionality Theorem, this transversal cuts sides

\overline{A C}

and

\overline{A E}

proportionally.

\frac{D E}{A D} = \frac{B C}{A B}

Substituting the length of

\overline{A B},

\overline{B C},

and

\overline{A D}

gives an equation which can be solved for the length of

\overline{D E} .

\frac{D E}{A D} = \frac{B C}{A B}

SubstituteValues

Substitute values

\frac{D E}{b} = \frac{a}{1}

DivByOne

$\frac{a}{1} = a$

\frac{D E}{b} = a

MultEqn

$LHS \cdot b = RHS \cdot b$

D E = a b

This construction gave a segment of length $a b .$

Discussion

Converse Triangle Proportionality Theorem

The converse of the Side-Splitter Theorem is also true.

If a segment is drawn between two sides of a triangle such that it divides the sides proportionally, the segment is parallel to the third side in the triangle.

Based on the diagram, the following relation holds true.

If $\frac{A D}{D C} = \frac{B E}{E C},$ then $\overline{D E} ∥ \overline{A B} .$

Proof

The given proportion can be rearranged to get the proportionality of two sides of

△ A B C

and

△ D E C .

\frac{A D}{D C} = \frac{B E}{E C}

▼

Rewrite

AddEqn

$LHS + 1 = RHS + 1$

\frac{A D}{D C} + 1 = \frac{B E}{E C} + 1

OneToFrac

Rewrite $1$ as $\frac{a}{a}$

\frac{A D}{D C} + \frac{D C}{D C} = \frac{B E}{E C} + \frac{E C}{E C}

AddFrac

Add fractions

\frac{A D + D C}{D C} = \frac{B E + E C}{E C}

Segment Addition Postulate

\frac{A C}{D C} = \frac{B C}{E C}

This means that

△ A B C

is a dilation of

△ D E C

from point

C

with scale factor

r = \frac{A C}{D C} = \frac{B C}{E C} .

A dilation moves a segment to a parallel segment, so the proof is complete.

If $\frac{A D}{D C} = \frac{B E}{E C},$ then $\overline{D E} ∥ \overline{A B} .$

Example

Solving Problems With the Converse Triangle Proportionality Theorem

Show that $P Q R S$ is a parallelogram.

Quadrilateral ABCD with points P,Q,R,S on the four sides. The given distances are AP=4.2, PB=1.8, BQ= 2.4, QC=5.6, CR=6.3, RD=2.7, DS=2.1 SA=4.8.

Hint

Draw the diagonals of quadrilateral $A B C D .$

Solution

Draw diagonal $\overline{A C}$ of quadrilateral $A B C D$ and focus on the two triangles $△ A B C$ and $△ A D C .$

The given measures of the segments make it possible to derive the ratios according to how the transversal

\overline{P Q}

divides sides

\overline{A B}

and

\overline{B C}

△ A B C .

\frac{P A}{P B} \frac{Q C}{Q B} = \frac{4 . 2}{1 . 8} = \frac{7}{3} = \frac{5 . 6}{2 . 4} = \frac{7}{3}

It can be seen that the two ratios are equal. Therefore, according to the converse of the Triangle Proportionality Theorem, the transversal

\overline{P Q}

is parallel to the diagonal

\overline{A C .}

A similar calculation shows that

\overline{S R}

cuts the sides of

△ A D C

proportionally. Therefore,

\overline{S R}

is also parallel to the diagonal

\overline{A C .}

\frac{S A}{S D} \frac{R C}{R D} = \frac{4 . 9}{2 . 1} = \frac{7}{3} = \frac{6 . 3}{2 . 7} = \frac{7}{3}

Since

\overline{P Q}

and

\overline{S R}

are both parallel to

\overline{A C,}

they are also parallel to each other.

After drawing the diagonal

\overline{B D},

the previously found proportions also show that

\overline{P S}

cuts the sides

\overline{A B}

and

\overline{A D}

△ A B D

proportionally. Additionally,

\overline{Q R}

cuts the sides

\overline{C B}

and

\overline{C D}

△ C B D

proportionally.

\frac{P A}{P B} \frac{Q C}{Q B} = \frac{S A}{S D} = \frac{7}{3} = \frac{R C}{R D} = \frac{7}{3}

This implies that

\overline{P S}

and

\overline{Q R}

are both parallel to

\overline{B D .}

Therefore, they are also parallel to each other.

This completes the proof that opposite sides of quadrilateral $P Q R S$ are parallel. Hence, by definition, it is a parallelogram.

Discussion

Three Parallel Lines Theorem

The following theorem is a corollary of the Side-Splitter Theorem.

If three parallel lines intersect two transversals, then they divide the transversals proportionally.

three parallel lines and two transversals

Applying the theorem to the diagram above, the following proportion can be written.

$\frac{U W}{W Y} = \frac{V X}{X Z}$

Proof

In the diagram, draw $\overline{V Y}$ and let $P$ be the point of intersection between this segment and line $s .$

Next, separate

△ U V Y

and

△ Y Z V .

Since

\overline{W P} ∥ \overline{U V},

by the Triangle Proportionality Theorem

\overline{W P}

divides

\overline{Y U}

and

\overline{Y V}

proportionally.

\frac{U W}{W Y} = \frac{V P}{P Y}

Applying the same reasoning in

△ Y Z V,

it can be said that

\overline{P X}

divides

\overline{V Y}

and

\overline{V Z}

proportionally.

\frac{V P}{P Y} = \frac{V X}{X Z}

Finally, the Transitive Property of Equality leads to the desired proportion, showing that the three parallel lines divide the transversals proportionally.

⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ \frac{U W}{W Y} = \frac{V P}{P Y} \frac{V P}{P Y} = \frac{V X}{X Z} \Rightarrow \frac{U W}{W Y} = \frac{V X}{X Z}

Pop Quiz

Practice Using the Three Parallel Lines Theorem

Example

Solving Problems Using the Three Parallel Lines Theorem

Construct points that divide the given segment into five congruent pieces.

Hint

Draw a different segment and extend it with four congruent copies.

Solution

Draw a ray starting at $A$ and use a compass to copy any length five times on this ray. This gives five points, $P_{1},$ $P_{2},$ $P_{3},$ $P_{4},$ and $P_{5} .$

Connect $B$ with the last point, $P_{5},$ and construct parallel lines to this segment through the other points. Mark the intersection points of these lines with segment $\overline{A B} .$

According to the Three Parallel Lines Theorem, these transversals divide segments $\overline{A B}$ and $\overline{A P_{5}}$ proportionally. Since, by construction, the segments on $\overline{A P_{5}}$ have equal length, this means that points $Q_{1},$ $Q_{2},$ $Q_{3},$ and $Q_{4}$ divide $\overline{A B}$ into congruent segments.

Explore

Investigating Inner Triangles of a Triangle

The examples until now were based on similar triangles generated by parallel lines. Is it possible to cut a triangle to two similar triangles using a line starting from a vertex?

Move the vertices and the point on the side of the triangle. It is possible to find an arrangement when the two inner triangles are similar to each other and to the original triangle. Find such a diagram.

Discussion

Right Triangle Similarity Theorem

The previous exploration can lead to the following claim.

Given a right triangle, if an altitude is drawn from the vertex of the right angle to the hypotenuse, then the two triangles formed are similar to the original triangle and to each other.

A right triangle with the altitude to the hypotenuse drawn

According to this theorem, there are three relations that hold true for the diagram above.

$△ C B D \sim △ A B C$
$△ A C D \sim △ A B C$
$△ C B D \sim △ A C D$

Proof

Start by showing separately the two triangles formed when the altitude

\overline{C D}

dissects

△ A B C .

By the Reflexive Property of Congruence,

∠ B ≅ ∠ B

and

∠ A ≅ ∠ A .

Also, since all right angles are congruent, it is obtained that

∠ B D C ≅ ∠ B C A

and

∠ C D A ≅ ∠ B C A .

$△ C B D$ and $△ A B C$	$△ A C D$ and $△ A B C$
$∠ B ≅ ∠ B$	$∠ A ≅ ∠ A$
$∠ B D C ≅ ∠ B C A$	$∠ C D A ≅ ∠ B C A$

Applying the Angle-Angle (AA) Similarity Theorem, it can be concluded that $△ C B D$ and $△ A B C$ are similar, and $△ A C D$ and $△ A B C$ are similar. Then, by the Transitive Property of Congruence, $△ A C D$ and $△ C B D$ are also similar.

△ C B D \sim △ A B C

and

△ A C D \sim △ A B C

⇓ △ C B D \sim △ A C D

Discussion

Corollaries of the Right Triangle Similarity Theorem

The following two claims are corollaries of the Right Triangle Similarity Theorem

Rule

Geometric Mean Altitude Theorem

Given a right triangle, if an altitude is drawn from the vertex of the right angle to the hypotenuse, then the measure of this altitude is the geometric mean between the measures of the two segments formed on the hypotenuse.

Right triangle with the altitude from the right angle drawn

Based on the diagram above and by definition of the geometric mean, the following relation holds true.

$C D^{2} = A D \cdot B D$ or $\frac{C D}{A D} = \frac{B D}{C D}$

The Geometric Mean Altitude Theorem is also known as the Right Triangle Altitude Theorem and the Geometric Mean Theorem.

Proof

According to the Right Triangle Similarity Theorem, the two triangles formed by the altitude

\overline{C D}

are similar.

△ C B D \sim △ A C D

Then, by definition of similar triangles, the lengths of corresponding sides are proportional.

$\frac{C D}{A D} = \frac{B D}{C D}$

Applying the Properties of Equality, this proportion can be rewritten without fractions.

$C D^{2} = A D \cdot B D$

Rule

Geometric Mean Leg Theorem

Given a right triangle, if the altitude is drawn from the vertex of the right angle to the hypotenuse, then the measure of each leg of the triangle is the geometric mean between the length of the hypotenuse and the length of the segment formed on the hypotenuse adjacent to the leg.

Based on the diagram above, the following relations hold true.

⎩ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎧ \frac{A C}{A D} = \frac{A B}{A C} \frac{C B}{D B} = \frac{A B}{C B} or {A C^{2} = A D \cdot A B C B^{2} = D B \cdot A B

Proof

According to the Right Triangle Similarity Theorem, the two triangles formed by the altitude

\overline{C D}

are similar to

△ A B C .

△ A B C \sim △ A C D △ A B C \sim △ C B D

This means that the corresponding parts of the triangles can be identified.

$△ A B C \sim △ A C D$	$△ A B C \sim △ C B D$
$\overline{A B} and \overline{A C}$ $\overline{A C} and \overline{A D}$ $\overline{C B} and \overline{C D}$	$\overline{A B} and \overline{C B}$ $\overline{A C} and \overline{C D}$ $\overline{C B} and \overline{D B}$

Then, by definition of similar triangles, the lengths of corresponding sides are proportional.

$△ A B C \sim △ A C D$	$△ A B C \sim △ C B D$
$\frac{A C}{A D} = \frac{A B}{A C}$	$\frac{C B}{D B} = \frac{A B}{C B}$

Note that for any two pairs of corresponding sides a similar proportion can be obtained. Now, applying the Properties of Equality, the proportion can be rewritten without fractions.

{A C^{2} = A D \cdot A B C B^{2} = D B \cdot A B

Example

Solving Problems With the Geometric Mean Leg Theorem

Represented on the figure $△ A B C$ is a right triangle and an altitude $\overline{C D}$ .

CD is an altitude of the right triangle ABC where D divides AB forming another right triangle ADC. AD is 28.8 cm and BC is 13 cm.

Using the given measurements, find the length of $\overline{C D} .$

Hint

First, find the length of $\overline{D B} .$

Solution

The answer can be found in two steps. The first step is to find the length of $\overline{D B} .$

Finding $D B$

The Geometric Mean Leg Theorem shows the connection between the length of $\overline{A B},$ $\overline{D B},$ and $\overline{C B} .$

$C B^{2} = D B \cdot A B$

The length of $\overline{A B}$ can be expressed using the length of $\overline{D B} .$

By substituting these expressions in the equation given by the Geometric Mean Leg Theorem, the results can be expressed in an equation that can then be solved for the length of

\overline{D B} .

C B^{2} = D B \cdot A B

SubstituteExpressions

Substitute expressions

1 3^{2} = x (28.8 + x)

▼

Rewrite

Distr

Distribute $x$

1 3^{2} = 28.8 x + x^{2}

SubEqn

$LHS - 1 3^{2} = RHS - 1 3^{2}$

0 = 28.8 x + x^{2} - 1 3^{2}

CalcPow

Calculate power

0 = 28.8 x + x^{2} - 169

RearrangeEqn

Rearrange equation

x^{2} + 28.8 x - 169 = 0

UseQuadForm

Use the Quadratic Formula: $a = 1, b = 28.8, c = - 169$

x = \frac{- 2 8 . 8 \pm 2 8 . 8 ^{2} - 4 ( 1 ) ( - 1 6 9 )}{2 ( 1 )}

▼

Evaluate right-hand side

CalcPowProd

Calculate power and product

x = \frac{- 2 8 . 8 \pm 8 2 9 . 4 4 + 6 7 6}{2}

AddTerms

Add terms

x = \frac{- 2 8 . 8 \pm 1 5 0 5 . 4 4}{2}

CalcRoot

Calculate root

x = \frac{- 2 8 . 8 \pm 3 8 . 8}{2}

WriteSumFrac

Write as a sum of fractions

x = - \frac{2 8 . 8}{2} \pm \frac{3 8 . 8}{2}

CalcQuot

Calculate quotient

x = - 14.4 \pm 19.4

This gives two solutions,

x = - 14.4 + 19.4 = 5,

and

x = - 14.4 - 19.4 = - 33.8 .

Since

x

represents the length of segment

\overline{D B},

only the positive solution is meaningful, as a length of a segment can not be negative.

D B = 5 cm

Finding $C D$

This gives the length of the other segment formed by the altitude in the right triangle $△ A B C .$

The Geometric Mean Altitude Theorem gives a connection between the length of $\overline{A D},$ $\overline{D B},$ and $\overline{C D} .$

$C D^{2} = A D \cdot D B$

This relationship can be used to find the length of the altitude.

C D^{2} = A D \cdot D B

SubstituteII

$A D = 28.8$ , $D B = 5$

C D^{2} = 28.8 (5)

▼

Solve for

C D

Multiply

C D^{2} = 144

CalcRoot

Calculate root

C D = 12

The length of the altitude

\overline{C D}

12

centimeters.

Pop Quiz

Practice the Geometric Mean Leg Theorem

Discussion

Pythagorean Theorem

For right triangles, the length of the hypotenuse squared equals the sum of the squares of the lengths of the legs.

Proof

Using Similarity

First, draw the altitude from the right angle to the hypotenuse. This divides the hypotenuse into two segments.

Next, apply the Geometric Mean Leg Theorem. Doing this relates the lengths of the legs to the length of the hypotenuse.

{a^{2} = x \cdot c b^{2} = y \cdot c

These two equations can be added. Then,

c

can be factored out from the right-hand side.

a^{2}^{+} b^{2} a^{2} + b^{2} a^{2} + b^{2} = x \cdot c = y \cdot c = x \cdot c + y \cdot c = (x + y) c

Drawing the altitude resulted in the outcome that

x + y

is equal to

c .

After substituting this into the equation above and simplifying, the Pythagorean Theorem is obtained.

a^{2} + b^{2} a^{2} + b^{2} = c \cdot c ⇕ = c^{2}

Closure

Solving Problems Using Triangle Similarity

To conclude this lesson, the opening challenge will be revisited. The challenge shows a diagram consisting of the Grim Reaper and two street lamps at $15$ and $10$ feet tall. How tall is the Grim Reaper?

Hint

The lamps, the head of the Grim Reaper, and the shadows of the Grim Reaper's head are on a straight line.

Solution

The lamps and the figure stand vertically. Hence, they can be represented by parallel segments $\overline{A B},$ $\overline{C D},$ and $\overline{E F}$ . Notice that the shadows just reach the lampposts. Taking a look at the shadow touching the taller post, it can be derived that the lamppost bottom $B,$ the head of the Grim Reaper $C,$ and the lamppost top $E$ are on a straight line. Applying this same logic to the other shadow implies that $A,$ $C,$ and $F$ are also on a straight line.

Segment $\overline{C D}$ splits both $△ A B F$ and $△ B E F .$ It is also parallel to one side of both triangles. That means the combination of two dilations maps $\overline{A B}$ to $\overline{E F} .$

A dilation with center $F$ and scale factor $D F : B F$ maps $\overline{A B}$ to $\overline{C D} .$
A dilation with center $B$ and scale factor $B F : B D$ maps $\overline{C D}$ to $\overline{E F} .$

The scale factor of this combined similarity transformation is the product of the two scale factors.

Continuing, notice that Point

D

is between

B

and

F .

Therefore, the Segment Addition Postulate guarantees that

B D + D F = B F .

Then, to divide this equality by

B F

and to rearrange it gives a relationship between the scale factors of the two dilations.

B D + D F = B F

▼

Rewrite

DivEqn

$LHS / B F = RHS / B F$

\frac{B D}{B F} + \frac{D F}{B F} = 1

SubEqn

$LHS - \frac{D F}{B F} = RHS - \frac{D F}{B F}$

\frac{B D}{B F} = 1 - \frac{D F}{B F}

$\frac{1}{LHS} = \frac{1}{RHS}$

\frac{B F}{B D} = \frac{1}{1 - \frac{D F}{B F}}

The product of the two scale factors is the scale factor of the similarity transformation that maps the

15 -

foot lamppost to the

10 -

foot lamppost. Hence, the scale factor is

10 : 15 .

An equation can now be written and solved to find the individual scale factors.

\frac{D F}{B F} \cdot \frac{B F}{B D} = \frac{1 0}{1 5}

SubstituteExpressions

Substitute expressions

\frac{D F}{B F} \cdot \frac{1}{1 - \frac{D F}{B F}} = \frac{1 0}{1 5}

▼

Solve for

\frac{D F}{B F}

Substitute

$\frac{D F}{B F} = x$

x \cdot \frac{1}{1 - x} = \frac{1 0}{1 5}

MultEqn

$LHS \cdot 15 (1 - x) = RHS \cdot 15 (1 - x)$

15 x = 10 (1 - x)

Distr

Distribute $10$

15 x = 10 - 10 x

AddEqn

$LHS + 10 x = RHS + 10 x$

25 x = 10

DivEqn

$LHS / 25 = RHS / 25$

x = \frac{1 0}{2 5}

CalcQuot

Calculate quotient

x = 0.4

Substitute

$x = \frac{D F}{B F}$

\frac{D F}{B F} = 0.4

This finding,

0.4,

is the scale factor of the dilation that maps

\overline{A B}

\overline{C D} .

Since

A B = 15,

the length of

\overline{C D}

can now be calculated.

C D = 0.4 (15) = 6

Segment

\overline{C D}

represents the Grim Reaper, who is now known to stand at

6

feet tall.

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Catch-Up and Review

Proof

Answer

Hint

Solution

Proof

Hint

Solution

Proof

Hint

Solution

Proof

Proof

Proof

Hint

Solution

Finding DB

Finding CD

Proof

Hint

Solution

Finding $D B$

Finding $C D$