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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.$△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 $

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 $

Substituting the length of $AB,$ $BC,$ and $AD$ gives an equation which can be solved for the length of $DE.$ $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.$$PBPA QBQC =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 $AC.$
$SDSA RDRC =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.
$PBPA QBQC =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.

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 $

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 =PYVP PYVP =XZVX ⇒WYUW =XZVX $

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, $P_{1},$ $P_{2},$ $P_{3},$ $P_{4},$ 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 $AB.$

According to the Three Parallel Lines Theorem, these transversals divide segments $AB$ and $AP_{5} $ proportionally. Since, by construction, the segments on $AP_{5} $ have equal length, this means that points $Q_{1},$ $Q_{2},$ $Q_{3},$ and $Q_{4}$ 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 $△ACD∼△ABC$

$⇓△CBD∼△ACD $

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.

$CD_{2}=AD⋅BD$ or $ADCD =CDBD $

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.

$△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.

$CD_{2}=AD⋅BD$

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 $△ABC.$

$△ABC∼△ACD△ABC∼△CBD $

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.

${AC_{2}=AD⋅ABCB_{2}=DB⋅AB $

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 $CB.$

$CB_{2}=DB⋅AB$

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.$$CB_{2}=DB⋅AB$

SubstituteExpressions

Substitute expressions

$13_{2}=x(28.8+x)$

Rewrite

Distr

Distribute $x$

$13_{2}=28.8x+x_{2}$

SubEqn

$LHS−13_{2}=RHS−13_{2}$

$0=28.8x+x_{2}−13_{2}$

CalcPow

Calculate power

$0=28.8x+x_{2}−169$

RearrangeEqn

Rearrange equation

$x_{2}+28.8x−169=0$

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 $CD.$

$CD_{2}=AD⋅DB$

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

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⋅cb_{2}=y⋅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⋅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.
$a_{2}+b_{2}a_{2}+b_{2} =c⋅c⇕=c_{2} $

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 $