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Similarity, Proof, and Trigonometry

Application of Congruence and Similarity Theorems

In this lesson, some interesting properties of quadrilaterals, trapezoids, and angle bisectors of triangles will be explored. Each of these will be proven using congruence and similarity.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Explore

Investigating Properties of a Quadrilateral

In the net, a quadrilateral, the segments divide the sides into eight congruent segments.

  • Use the measuring tool to investigate how the segments divide each other inside the quadrilateral.
  • Explore what happens when the vertices are moved!

Example

Investigating the Midpoints of a Quadrilateral

The Triangle Midsegment Theorem gives a relationship between a midsegment and a side of a triangle. There too, is an exciting result for quadrilaterals, formed by the midpoints of the sides of a quadrilateral. Illustrated in the diagram are P, Q, R, and S which are midpoints of the sides of the quadrilateral ABCD.

Show that PQRS is a parallelogram, and that PR and QS bisect each other.

Hint

Draw a diagonal in quadrilateral ABCD and focus on the two triangles.

Solution

Parallelogram

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

According to the Triangle Midsegment Theorem, both PQ and SR are parallel to the diagonal AC, and they are half the length of AC. That means these midsegments are parallel to each other, and they have the same length.
These relationships can be plotted on the diagram.

Similarly, PS and QR are also parallel and have the same length.

By definition, when the opposite sides of a quadrilateral are parallel, then it is a parallelogram. Therefore, the quadrilateral PQRS is a parallelogram.

Bisecting Diagonals

To show that the diagonals PR and QS bisect each other, focus on two of the triangles formed by these diagonals.

These triangles contain the following properties.

Claim Justification
PQSR Proved previously
RPQPRS Alternate Interior Angles Theorem
PQSRSQ Alternate Interior Angles Theorem

These claims can be shown in the diagram.

It can be seen that triangles PQM and RSM have two pairs of congruent angles, and the included sides are also congruent. According to the Angle-Side-Angle (ASA) Congruence Theorem, the triangles are congruent.
Corresponding parts of congruent triangles are congruent.
This completes the proof that PR and QS bisect each other.

Example

Investigating Properties of a General Trapezoid

The following example discusses a property of a general trapezoid. On the diagram ABCD is a trapezoid and EF is parallel to the bases through M, the intersection of the diagonals.

Show that M is the midpoint of EF.

Hint

Look for similar triangles.

Solution

There are several pairs of similar triangles on the diagram. Using the scale factors of the similarity transformations between these triangles, the length of EM and MF can be expressed in terms of the length of the bases AB and DF. Here is the outline of a possible approach.

  • Step 1: Investigate triangles ABM and CDM.
  • Step 2: Investigate triangles DEM and DAB to express the length of EM in terms of the length of the bases AB and DC.
  • Step 3: Investigate triangles CMF and CAB to express the length of MF in terms of the length of the bases AB and DC.
  • Step 4: Compare the expressions for the length of EM and the length of MF.

Here are the details.

Step 1

Focus on the triangles formed by the bases and the diagonals of the trapezoid.

The following table contains some information about these triangles.

Claim Justification
BACDCA Alternate Interior Angles Theorem
ABDCDB Alternate Interior Angles Theorem

This can be indicated on the diagram.

According to the Angle-Angle (AA) Similarity Theorem, this means that the two triangles are similar, so the corresponding sides are proportional.

Step 2

Focus now on the left side of the trapezoid.

Since EM is parallel to AB, a dilated image of DEM is DAB. The scale factor can be written in two different ways.
In this equality DB can be replaced by DM+MB.
According to Step 1, the ratio MB:DM is the same as the ratio AB:DC. Substituting this in the equation above gives an equation that can be solved for EM.
Solve for EM

This calculation gave an expression for the length of EM in terms of the lengths of the bases of the trapezoid.

Step 3

On the right of the trapezoid there are two more similar triangles, ABC and MFC.

The scale factor of the dilation between these two triangles can be written two different ways.
As in Step 2, this equation can also be solved for MF. In this case the second equation of the result of Step 1 can be used.
Simplify right-hand side

Solve for MF

Step 4

The results of Step 2 and Step 3 show that the expressions for EM and MF are identical, so these two segments are congruent.
This completes the proof that M is the midpoint of EF. Move the point on the base of the trapezoid to see an illustration of the claim.

Explore

Investigating Angle Bisectors of Triangles

The next part of this lesson focuses on triangles. The diagram shows a triangle with one of its angle bisectors drawn. Move the vertices of the triangle and find a relationship between the displayed segment measures.

Discussion

Triangle Angle Bisector Theorem

The relationship stated in the following theorem can be checked on the previous applet for different triangles.

The angle bisector of an interior angle of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.

triangle and angle bisector

In the figure, if is an angle bisector, then the following equation holds true.

Proof

In ABC, consider the angle bisector that divides ABC into two congruent angles. Let ∠1 and ∠2 be these congruent angles.

triangle and angle bisector

By the Parallel Postulate, a parallel line to can be drawn through A. Additionally, if BC is extended, it will intersect this line. Let E be their point of intersection.

triangle and point of intersection of the lines

Let ∠3 be the alternate interior angle to ∠1 formed at A. Also, let ∠4 be the corresponding angle to ∠2 formed at E.

triangle and the pair of corresponding angles and alternate angles
By the Corresponding Angles Theorem, ∠2 is congruent to ∠4. Remember that it is also known that ∠1 is congruent to ∠2. By the Transitive Property of Congruence, ∠1 and ∠4 are congruent angles.
Additionally, by the Alternate Interior Angles Theorem, ∠1 is congruent to ∠3. Using the Transitive Property of Congruence one more time, it can be said that ∠3 and ∠4 are also congruent angles.
This can be shown in the diagram.
triangle and congruent angles
Note that ACE is divided by which is parallel to AE. Therefore, by the Triangle Proportionality Theorem, divides the other two sides of this triangle proportionally.
The Converse Isosceles Triangle Theorem states that if two angles in a triangle are congruent, the sides opposite them are congruent. This means that EB is congruent to AB. Therefore, by the definition of congruent segments, they have the same length. AB can be substituted for EB in the above proportion.

Pop Quiz

Practice the Triangle Angle Bisector Theorem

Find the measurement of the segment as indicated in the applet.

Example

Solving Problems With the Triangle Angle Bisector Theorem

In ABC, segment AD is the angle bisector of the right angle at A, and DE is perpendicular to AC. The length of the legs AB and AC are 5 and 12, respectively.

Find the length of AE. Write the answer in exact form as a fraction.

Hint

Start with finding the length of the hypotenuse and the length of CD.

Solution

Mark the lengths which were given in the prompt onto the diagram.

The length of the hypotenuse of the triangle can be found using the Pythagorean Theorem.
As can be seen on the diagram, the length of BC is the sum of BD and DC. When rearranged, this can be written as BD=BCDC. Furthermore, let x represent DC, as it is unknown. Then, since BC was just found, it can be said that BD=13x.
According to the Angle Bisector Theorem, an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle. In this case, that would indicate proportionality between the ratio of the two segments of the hypotenuse and the ratio of the altitude and base.
Substituting the expressions from the diagram gives an equation that can be solved for x, which represents the length of DC.
Solve for x
12(13x)=5x
15612x=5x
156=17x
This gives the lengths of the segments on the hypotenuse.
Recall that the task is to find the length of AE. Using similar logic as before, if z is used to represent the length of AE, then EC=12z.
Since both BA and DE are perpendicular to AC, these segments are parallel. According to the Triangle Proportionality Theorem, this means that DE divides sides BC and AC proportionally.
Substituting the expressions from the diagram gives an equation that can be solved for z, which represents the length of AE.
Solve for z
65(12z)=156z
78065z=156z
780=221z
The length of AE is

Discussion

Converse Triangle Angle Bisector Theorem

According to the Angle Bisector Theorem, an angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle. The converse of this statement is also true.

If a segment from a vertex B of a triangle divides the opposite side in proportion to the sides meeting at B, then the segment is an angle bisector of the triangle.

triagle and angle bisector

Based on the figure, the following conditional statement holds true.

This theorem is the converse of the Triangle Angle Bisector Theorem.

Proof

Consider ABC and the segment that connects vertex B with its opposite side. Let D be the point of intersection of the segment from B and AC. Now, CB will be extended to a point E such that BE equals AB. Additionally, a segment from A to E will be constructed.

triangle and the point of intersection of the lines
It is given that BD divides the opposite side in proportion to the sides meeting at B.
Because BE is equal to AB, by the Substitution Property of Equality BE can be substituted for AB in the proportion.
Therefore, BD is a segment between two sides of ACE that divides EC and AC proportionally. Then, by the Converse Triangle Proportionality Theorem it can be stated that EA is parallel to BD.
triangle, parallel segments and congruent segments
It is seen that AEB and DBC are corresponding angles. By the Corresponding Angles Theorem, AEB is congruent to DBC. Furthermore, EAB and ABD are alternate interior angles, and by the Alternate Interior Angles Theorem these two angles are also congruent.
Because BE=AB, by the Isosceles Triangle Theorem AEB is congruent to EAB.
triangle, parallel segments, congruent segments, and congruent angles
Since CBD and EAB are both congruent to AEB, by the Transitive Property of Congruence, it follows that CBD and EAB are congruent angles.
By the same property, since ABD and CBD are both congruent to EAB, they are congruent angles.

Therefore, by the definition of an angle bisector BD is an angle bisector of the triangle.

Example

Solving Problems With the Converse Triangle Angle Bisector Theorem

On the diagram, the markers on line are equidistant, the circles are centered at A and at B, and C is the point of intersection of the circles.

Point A is at the third marker, the circle centered at A passes through the seventh marker. Point B is at the eights marker, the circle centered at B passes through the second marker. Point D is at the fifth marker.

Show that CD bisects ACB.

Hint

Express the lengths of the line segments in terms of the distance between consecutive markers.

Solution

The lengths of some line segments can be expressed in terms of the distance between consecutive markers.

Claim Justification
AD=2 By counting the markers
DB=3 By counting the markers
AC=4 Segment AC is a radius of the circle centered at A. Counting markers shows that the radius of this circle is 4 units long.
CB=6 Segment BC is a radius of the circle centered at B. Counting markers shows that the radius of this circle is 6 units long.

These measurements can be indicated on the diagram.

The ratio of two sides of the triangle can be simplified.
That equals the ratio of the two segments on the third side of the triangle.
According to the converse of the Angle Bisector Theorem, this relationship between the proportions means that CD bisects ACB.

Closure

Analyzing Properties of a Quadrilateral

At the beginning of this lesson, the following net was investigated. Recall that the segments drawn inside the net, a quadrilateral, cut the sides into congruent parts.
Show that all segments are cut by the others into congruent parts.

Hint

Use the knowledge that the segments connecting the midpoints of opposite sides of any quadrilateral bisect each other.

Solution

In the first exercise of this lesson, it was proved that the segments connecting the midpoints of opposite sides of any quadrilateral bisect each other.
This statement can be used several times considering different quadrilaterals to prove the claim that all segments are cut by the others into congruent parts.

Step 1

First, consider only the midpoints of the original quadrilateral and the segments connecting these midpoints, They intersect at the mark.

As it can be seen, these segments bisect each other.

Step 2

The segments connecting the midpoints of opposite sides cut the original quadrilateral into two smaller quadrilaterals. Focus on the segments connecting the midpoints of opposite sides of this smaller quadrilateral.

As shown, these segments also bisect each other.

Step 3

Next, focus on another quadrilateral that differs from the previous two smaller ones. Again, take note of the segments connecting the midpoints of opposite sides.

These segments also bisect each other.

Step 4

Now, consider a quarter of the original quadrilateral. Mark the segments that connect the midpoints of its opposite sides.

Again, these bisect each other.

Step 5

Using similar logic as in the previous steps, intersection points in similar positions can also be marked.
This shows that when considering every second segment in both directions, the segments cut each other into congruent pieces.

Step 6

The other intersection points can also be found as midpoints of certain segments. Therefore, continuing this process shows that all segments are cut by the others into congruent pieces.

Extra

The solution above was based on finding midpoints again and again. Similar arguments can be used to prove the claim for 2, 4, 8, , segments as well. The following applet can be used to check that a similar statement is also true when the number of segments on the sides is not a power of 2.
While this claim will not be proved here, it is a worthwhile concept to consider.
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