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.

Based on the diagram, the following relation holds true.

If DE∥AB, then DCAD=ECBE

### Proof

Since DE and AB are parallel, by the Corresponding Angles Theorem, ∠CDE and ∠CAB are congruent. Similarly, ∠CED and ∠CBA are congruent.

Therefore, by the Angle-Angle Similarity Theorem, △ABC and △DEC are similar. Consequently, their corresponding sides are proportional.

△ABC∼△DEC⇓DCAC=ECBC

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

ACBC=AD+DC=BE+EC

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

DCAC=ECBC

AC=AD+DC, BC=BE+EC

DCAD+DC=ECBE+EC

Write as a sum of fractions

DCAD+DCDC=ECBE+ECEC

Simplify quotient

DCAD+1=ECBE+1

LHS−1=RHS−1

DCAD=ECBE

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 ab.

### Answer

### Hint

Use that ab: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.

In △ACE the transversal BD is parallel to the side CE. According to the Triangle Proportionality Theorem, this transversal cuts sides AC and AE proportionally.

ADDE=ABBC

Substituting the length of AB, BC, and AD gives an equation which can be solved for the length of DE.

This construction gave a segment of length ab.

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 DCAD=ECBE, then DE∥AB.

### Proof

The given proportion can be rearranged to get the proportionality of two sides of △ABC and △DEC.

DCAD=ECBE

DCAC=ECBC

This means that △ABC is a dilation of △DEC from point C with scale factor r=DCAC=ECBC.A dilation moves a segment to a parallel segment, so the proof is complete.

If DCAD=ECBE, then DE∥AB.

Example

## Solving Problems With the Converse Triangle Proportionality Theorem

Show that PQRS is a parallelogram.

### Hint

Draw the diagonals of quadrilateral ABCD.

### Solution

Draw diagonal AC of quadrilateral ABCD and focus on the two triangles △ABC and △ADC.

The given measures of the segments make it possible to derive the ratios according to how the transversal PQ divides sides AB and BC of △ABC.

PBPAQBQC=1.84.2=37=2.45.6=37

It can be seen that the two ratios are equal. Therefore, according to the converse of the Triangle Proportionality Theorem, the transversal PQ is parallel to the diagonal AC. A similar calculation shows that SR cuts the sides of △ADC proportionally. Therefore, SR is also parallel to the diagonal AC.

SDSARDRC=2.14.9=37=2.76.3=37

Since PQ and SR are both parallel to AC, they are also parallel to each other.

After drawing the diagonal BD, the previously found proportions also show that PS cuts the sides AB and AD of △ABD proportionally. Additionally, QR cuts the sides CB and CD of △CBD proportionally.

PBPAQBQC=SDSA=37=RDRC=37

This implies that PS and QR are both parallel to BD. Therefore, they are also parallel to each other.

This completes the proof that opposite sides of quadrilateral PQRS 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.

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

WYUW=XZVX

### Proof

In the diagram, draw VY and let P be the point of intersection between this segment and line s.

Next, separate △UVY and △YZV.

Since WP∥UV, by the Triangle Proportionality Theorem WP divides YU and YV proportionally.

WYUW=PYVP

Applying the same reasoning in △YZV, it can be said that PX divides VY and VZ proportionally.

PYVP=XZVX

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

⎩⎪⎪⎪⎪⎨⎪⎪⎪⎪⎧WYUW=PYVPPYVP=XZVX⇒WYUW=XZVX

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, P1, P2, P3, P4, and P5.

Connect B with the last point, P5, and construct parallel lines to this segment through the other points. Mark the intersection points of these lines with segment AB.

According to the Three Parallel Lines Theorem, these transversals divide segments AB and AP5 proportionally. Since, by construction, the segments on AP5 have equal length, this means that points Q1, Q2, Q3, and Q4 divide AB 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.

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

- △CBD∼△ABC
- △ACD∼△ABC
- △CBD∼△ACD

### Proof

Start by showing separately the two triangles formed when the altitude CD dissects △ABC.

By the Reflexive Property of Congruence, ∠B≅∠B and ∠A≅∠A. Also, since all right angles are congruent, it is obtained that ∠BDC≅∠BCA and ∠CDA≅∠BCA.

△CBD and △ABC | △ACD and △ABC |
---|---|

∠B≅∠B | ∠A≅∠A |

∠BDC≅∠BCA | ∠CDA≅∠BCA |

Applying the Angle-Angle (AA) Similarity Theorem, it can be concluded that △CBD and △ABC are similar, and △ACD and △ABC are similar. Then, by the Transitive Property of Congruence, △ACD and △CBD are also similar.

△CBD∼△ABC and △ACD∼△ABC

⇓△CBD∼△ACD

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.

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

CD2=AD⋅BD or ADCD=CDBD

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 CD are similar.

△CBD∼△ACD

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

ADCD=CDBD

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

CD2=AD⋅BD

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.

⎩⎪⎪⎪⎨⎪⎪⎪⎧ADAC=ACABDBCB=CBABor{AC2=AD⋅ABCB2=DB⋅AB

### Proof

According to the Right Triangle Similarity Theorem, the two triangles formed by the altitude CD are similar to △ABC.

△ABC∼△ACD△ABC∼△CBD

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

△ABC∼△ACD | △ABC∼△CBD |
---|---|

ABandAC ACandAD | ABandCB ACandCD |

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

△ABC∼△ACD | △ABC∼△CBD |
---|---|

ADAC=ACAB | DBCB=CBAB |

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.

{AC2=AD⋅ABCB2=DB⋅AB

Example

## Solving Problems With the Geometric Mean Leg Theorem

Represented on the figure △ABC is a right triangle and an altitude CD.

Using the given measurements, find the length of CD.

### Hint

First, find the length of DB.

### Solution

The answer can be found in two steps. The first step is to find the length of DB.

### Finding DB

The Geometric Mean Leg Theorem shows the connection between the length of AB, DB, and CB.

CB2=DB⋅AB

The length of AB can be expressed using the length of DB.

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 DB.

CB2=DB⋅AB

SubstituteExpressions

Substitute expressions

132=x(28.8+x)

▼

Rewrite

Distribute x

132=28.8x+x2

LHS−132=RHS−132

0=28.8x+x2−132

CalcPow

Calculate power

0=28.8x+x2−169

Rearrange equation

x2+28.8x−169=0

UseQuadForm

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

x=2(1)-28.8±28.82−4(1)(-169)

▼

Evaluate right-hand side

CalcPowProd

Calculate power and product

x=2-28.8±829.44+676

AddTerms

Add terms

x=2-28.8±1505.44

Calculate root

x=2-28.8±38.8

Write as a sum of fractions

x=-228.8±238.8

Calculate quotient

x=-14.4±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 DB, only the positive solution is meaningful, as a length of a segment can not be negative.

DB=5cm

### Finding CD

This gives the length of the other segment formed by the altitude in the right triangle △ABC.

The Geometric Mean Altitude Theorem gives a connection between the length of AD, DB, and CD.

CD2=AD⋅DB

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

CD2=AD⋅DB

AD=28.8, DB=5

CD2=28.8(5)

CD=12

The length of the altitude CD is 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 SimilarityFirst, 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.

{a2=x⋅cb2=y⋅c

These two equations can be added. Then, c can be factored out from the right-hand side.

a2+b2a2+b2a2+b2=x⋅c=y⋅c=x⋅c+y⋅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.

a2+b2a2+b2=c⋅c⇕=c2

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 AB, CD, and EF. 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 CD splits both △ABF and △BEF. It is also parallel to one side of both triangles. That means the combination of two dilations maps AB to EF.

- A dilation with center F and scale factor DF:BF maps AB to CD.
- A dilation with center B and scale factor BF:BD maps CD to EF.

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 BD+DF=BF. Then, to divide this equality by BF and to rearrange it gives a relationship between the scale factors of the two dilations.

BD+DF=BF

▼

Rewrite

BDBF=1−BFDF1

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.

BFDF⋅BDBF=1510

SubstituteExpressions

Substitute expressions

BFDF⋅1−BFDF1=1510

BFDF=0.4

This finding, 0.4, is the scale factor of the dilation that maps AB to CD. Since AB=15, the length of CD can now be calculated.

CD=0.4(15)=6

Segment CD represents the Grim Reaper, who is now known to stand at 6 feet tall.