4. Similarity Theorems About Triangles
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Chapter 2
4.
Similarity Theorems About Triangles
Triangles hold a special place, especially when it comes to their proportions. The lesson delves into the intricacies of right triangle proportionality theorems, shedding light on how triangles can be similar based on certain conditions. One of the key takeaways is the Triangle Proportionality Theorem, which states that if a segment divides two sides of a triangle proportionally, it runs parallel to the third side. This theorem, along with others like the Geometric Mean Altitude Theorem, provides a deeper understanding of the relationships between different segments of a triangle. Such knowledge is not just theoretical; it has practical implications, helping in solving real-world problems that involve triangles and their proportions.
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Lesson Settings & Tools
| | 18 Theory slides |
| | 10 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Similarity Theorems About Triangles
Slide of 18
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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Challenge
Solving Problems Using Triangle Similarity
The street lamps are 15 and 10 feet tall. How tall is the Grim Reaper?
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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!
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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 AD/DC=BE/EC
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 ⇓ AC/DC=BC/EC Applying the Segment Addition Postulate, both numerators can be rewritten. AC &= AD+DC BC &= BE+EC Substituting these expressions into the equation above, the required proportion will be obtained.
AC/DC=BC/EC
SubstituteII
AC= AD+DC, BC= BE+EC
AD+DC/DC=BE+EC/EC
WriteSumFrac
Write as a sum of fractions
AD/DC+DC/DC=BE/EC+EC/EC
SimpQuot
Simplify quotient
AD/DC+1=BE/EC+1
SubEqn
LHS-1=RHS-1
AD/DC=BE/EC
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Pop Quiz
Practice Using the Triangle Proportionality Theorem
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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. DE/AD=BC/AB Substituting the length of AB, BC, and AD gives an equation which can be solved for the length of DE.
DE/AD=BC/AB
SubstituteValues
Substitute values
DE/b=a/1
DivByOne
a/1=a
DE/b=a
MultEqn
LHS * b=RHS* b
DE=ab
This construction gave a segment of length ab.
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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 AD/DC=BE/EC, then DE ∥ AB.
Proof
The given proportion can be rearranged to get the proportionality of two sides of △ ABC and △ DEC.
AD/DC=BE/EC
AC/DC=BC/EC
This means that △ ABC is a dilation of △ DEC from point C with scale factor r = ACDC= BCEC. A dilation moves a segment to a parallel segment, so the proof is complete.
If AD/DC=BE/EC, then DE ∥ AB.
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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. PA/PB&=4.2/1.8=7/3 [0.25cm] QC/QB&=5.6/2.4=7/3 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. SA/SD&=4.9/2.1=7/3 [0.25cm] RC/RD&=6.3/2.7=7/3 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. PA/PB&=SA/SD=7/3 [0.25cm] QC/QB&=RC/RD=7/3 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.
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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.
UW/WY = VX/XZ
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. UW/WY = VP/PY Applying the same reasoning in △ YZV, it can be said that PX divides VY and VZ proportionally. VP/PY = VX/XZ Finally, the Transitive Property of Equality leads to the desired proportion, showing that the three parallel lines divide the transversals proportionally.
UW/WY = VP/PY VP/PY = VX/XZ ⇒ UW/WY = VX/XZ
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Practice Using the Three Parallel Lines Theorem
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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 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.
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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.
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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
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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 given diagram and by the definition of the geometric mean, the following relation holds true.
CD^2=AD* BD or CD/AD = BD/CD
Other names for the Geometric Mean Altitude Theorem are 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.
CD/AD = BD/CD
Applying the Properties of Equality, this proportion can be rewritten without fractions.
CD^2=AD* BD
Rule
Geometric Mean Leg Theorem
In 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 given triangle, where the altitude CD is drawn from the vertex of the right triangle at C to the hypotenuse at D, the following relations hold true.
AC/AD = AB/AC [0.4cm] CB/DB = AB/CB or AC^2 = AD * AB CB^2 = 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 |
|---|---|
| AB and AC AC and AD |
AB and CB AC and CD |
Then, by definition of similar triangles, the lengths of corresponding sides are proportional.
| △ ABC ~ △ ACD | △ ABC ~ △ CBD |
|---|---|
| AC/AD = AB/AC | CB/DB = AB/CB |
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.
AC^2 = AD * AB CB^2 = DB * AB
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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.
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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.
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=- 28.8 ± sqrt(28.8^2-4( 1)( - 169))/2( 1)
▼
Evaluate right-hand side
CalcPowProd
Calculate power and product
x = - 28.8 ± sqrt(829.44 + 676)/2
AddTerms
Add terms
x = - 28.8 ± sqrt(1505.44)/2
CalcRoot
Calculate root
x = - 28.8 ± 38.8/2
WriteSumFrac
Write as a sum of fractions
x=-28.8/2±38.8/2
CalcQuot
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.
CD^2=AD* DB
This relationship can be used to find the length of the altitude.
The length of the altitude CD is 12 centimeters.
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Practice the Geometric Mean Leg Theorem
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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. a^2 = x* c b^2 = y* c These two equations can be added. Then, c can be factored out from the right-hand side. a^2 &= x* c [-0.15cm] ^+ b^2 &= y* c [-0.5ex] [-3ex] a^2+b^2 &= x* c+ y* c a^2+b^2 &=( 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 &= c* c &⇕ a^2+b^2 &= c^2
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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?
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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
BF/BD=1/1- DFBF
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.
DF/BF*BF/BD=10/15
SubstituteExpressions
Substitute expressions
DF/BF*1/1- DFBF=10/15
▼
Solve for DF/BF
Substitute
DF/BF= x
x*1/1- x=10/15
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=10/25
CalcQuot
Calculate quotient
x=0.4
Substitute
x= DF/BF
DF/BF=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.
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Similarity Theorems About Triangles
Exercise 2.1
Just before the holidays, Ignacio bought a new house with a few friends. He wants to determine the height of the house. To do this, he places a Christmas tree some distance away from the house. He positions himself in front of the tree such that from his perspective, the top of the house and the top of the Christmas tree coincide. He then made his measurements.
How high is the house? Round the answer to one decimal place.
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Examining the diagram, we can draw the following pair of similar right triangles.
Using the information that the triangles are similar, we can write a proportionality. h/1.10=3.10+13.2/3.10 Let's solve this equation for h.
To determine the full height of the house, we must add the height of Ignacio to h. 1.7+5.78387...≈ 7.5 m The height of the house is about 7.5 meters.
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Solution
Magdalena starts riding her bike from A at the same time that Ali begins riding his bike from B. Both are riding towards E, their meeting place.
After 20 minutes, Ali reaches D. After another 10 minutes he arrives at E. Assuming that Ali rides with a constant speed, what is Magdalena's speed in miles per hour if she arrives at E at the same time as Ali?
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To determine speed, we need to know distance and time. For Ali, we have both of these measures. We know that to travel BD takes Ali 20 minutes, and traveling AC takes Ali 10 minutes — half the time it takes to travel BD. Since his speed is constant, the distance DE must be half the distance of BD.
Notice that CD is parallel to AB. This means we can use the Triangle Proportionality Theorem to find the distance of CE. CE/4=2.5/5 ⇔ CE=2 Let's express this on a diagram.
The segments can be added to learn that Magdalena has to bike 6 miles in total. Like Ali, she arrives at E after 30 minutes, which can be expressed as 0.5 hours. With this information, we can determine her speed in miles per hour.
Magdalena's speed is 12 miles per hour.
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Solution
Consider the following diagram.
Determine the length of ED such that the area of △ DBE is 50 % of △ ABC.
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Notice that both △ DBE and △ ABC are right triangles. Therefore, to determine their area we must find the length of their legs. To do so, we will introduce a couple of variables. For △ DBE we will label the vertical leg as x and the horizontal leg as y.
Let's keep in mind that we are looking for the length of ED, which means we want to find x.
Similar Triangles
The right triangles share one acute angle. Therefore, by the Angle-Angle Similarity Theorem, the triangles are similar. For similar polygons it is true that the ratio of any pair of corresponding sides is the same. y/12=x/16 With this equation we can write y in terms of x. y/12=x/16 ⇔ y=3x/4
Area △ ABC
We know the base and height of △ ABC. That is enough information to determine its area.
The area of △ ABC is 96 square inches.
Area △ DBE
By using the expressions for the legs of △ DBE we can write an expression for its area.
Relation Between the Areas of the Triangles
We know that the area of the smaller triangle is half that of the bigger triangle. Let's use this information to write an expression. A_(△ DBE)=0.5(A_(△ ABC)) ⇓ 3x^2/8=0.5(96) By solving this equation for x we find ED.
When ED equals 8sqrt(2) inches, △ DBE has an area that is 50 % of △ ABC.
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Solution
Determine the value of x and y.
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We will begin by finding the value of x. Notice that EG, DF, and BA are all perpendicular to CA. This means EG∥DF∥BA. Since the segments are parallel, we can use the Three Parallel Lines Theorem to write a proportionality containing x and y. x/6 =y/2y Let's solve this equation for x.
As we can see, x equals 3. Let's add this information to the diagram.
Notice that △ ABC is a right triangle where we have expressions for both legs and the hypotenuse. Vertical Leg:& 3+3+6=12 Horizontal Leg:& 2y+6 Hypotenuse:& 2y+y+y=4y This means we can use the Pythagorean Theorem to write an equation containing y.
Let's solve for y by completing the square.
For the given triangle x=3 and y=5.
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Solution
Use a compass to find at what point AB is divided in a ratio of 2:1?
{"type":"choice","form":{"alts":["X","Y","Z","R","S","T","U"],"noSort":true},"formTextBefore":"","formTextAfter":"","answer":5}
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To divide AB in a ratio of 2:1, there are 6 steps. For the first 3 steps, we have to draw a segment at an angle from AB. Then we will divide this segment into three congruent pieces and connect the endpoint of the third piece with the endpoint of AB.
To continue, we need to copy ∠ AEB. This will allow us to draw parallel segments to EB from C and D that intersects AB.
Since AC, CD, and DE are of the same length, the parallel segments to EB divides AB into three congruent segments. If we add back the seven possible points to the segment, we can identify the correct option.
As we can see, T is the correct option.
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Solution
Similarity Theorems About Triangles
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