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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?

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

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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!

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

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.$DCAC =ECBC $

SubstituteII

AC=AD+DC, BC=BE+EC

$DCAD+DC =ECBE+EC $

WriteSumFrac

Write as a sum of fractions

$DCAD +DCDC =ECBE +ECEC $

SimpQuot

Simplify quotient

$DCAD +1=ECBE +1$

SubEqn

LHS−1=RHS−1

$DCAD =ECBE $

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.

Use that ab:b=a:1.

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 $

SubstituteValues

Substitute values

$bDE =1a $

DivByOne

$1a =a$

$bDE =a$

MultEqn

LHS⋅b=RHS⋅b

DE=ab

This construction gave a segment of length ab.

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.

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

$DCAD =ECBE $

$DCAC =ECBC $

If $DCAD =ECBE ,$ then DE∥AB.

Show that PQRS is a parallelogram.

Draw the diagonals of quadrilateral ABCD.

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.This completes the proof that opposite sides of quadrilateral PQRS are parallel. Hence, by definition, it is a parallelogram.

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.

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.

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

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

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.

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.

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

Start by separating the two triangles formed by the altitude CD from △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

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

or CD2=AD⋅BD

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

According to the Right Triangle Similarity Theorem, the two triangles formed by the altitude $CD$ are similar.
Then, by definition of similar triangles, the lengths of corresponding sides are proportional.

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

CD2=AD⋅BD

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 =ACAB DBCB =CBAB or{AC_{2}=AD⋅ABCB_{2}=DB⋅AB $

According to the Right Triangle Similarity Theorem, the two triangles formed by the altitude CD are similar to △ABC.
Then, by definition of similar triangles, the length of corresponding sides are proportional.

△ABC∼△ACD | △ABC∼△CBD |
---|---|

$ADAC =ACAB $ | $DBCB =CBAB $ |

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

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

Using the given measurements, find the length of CD.

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First, find the length of DB.

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

Distr

Distribute x

132=28.8x+x2

SubEqn

LHS−132=RHS−132

0=28.8x+x2−132

CalcPow

Calculate power

0=28.8x+x2−169

RearrangeEqn

Rearrange equation

x2+28.8x−169=0

UseQuadForm

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

$x=2(1)-28.8±28.8_{2}−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 $

CalcRoot

Calculate root

$x=2-28.8±38.8 $

WriteSumFrac

Write as a sum of fractions

$x=-228.8 ±238.8 $

CalcQuot

Calculate quotient

x=-14.4±19.4

$DB=5cm $

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

In a right triangle, the length of the hypotenuse squared equals the sum of the squares of the lengths of the legs.

First, the altitude from the right angle to the hypotenuse will be drawn. This will divide the hypotenuse into two segments.

By applying the Geometric Mean Leg Theorem, the lengths of the legs can be related to the length of the hypotenuse.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?

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

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.

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

$BDBF =1−BFDF 1 $

$BFDF ⋅BDBF =1510 $

SubstituteExpressions

Substitute expressions

$BFDF ⋅1−BFDF 1 =1510 $

Solve for $BFDF $

Substitute

$BFDF =x$

$x⋅1−x1 =1510 $

MultEqn

LHS⋅15(1−x)=RHS⋅15(1−x)

15x=10(1−x)

Distr

Distribute 10

15x=10−10x

AddEqn

LHS+10x=RHS+10x

25x=10

DivEqn

$LHS/25=RHS/25$

$x=2510 $

CalcQuot

Calculate quotient

x=0.4

Substitute

$x=BFDF $

$BFDF =0.4$

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