Proportionality Theorems for Triangles

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!

Pop Quiz

Practice Using the Triangle Proportionality Theorem

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.

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.

Discussion

Right Triangle Similarity Theorem

The previous exploration can lead to the following claim.

Discussion

Corollaries of the Right Triangle Similarity Theorem

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

Pop Quiz

Practice the Geometric Mean Leg Theorem

Discussion

Pythagorean Theorem

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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