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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.**

- Properties of congruence.
- Properties of similarity.

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!

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.

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

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.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.

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 |
---|---|

PQ≅SR | Proved previously |

∠RPQ≅∠PRS | Alternate Interior Angles Theorem |

∠PQS≅∠RSQ | 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.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.

Look for similar triangles.

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.

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 |
---|---|

∠BAC≅∠DCA | Alternate Interior Angles Theorem |

∠ABD≅∠CDB | 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.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.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.$MFAB =MCAC $

Substitute

AC=AM+MC

$MFAB =MCAM+MC $

Simplify right-hand side

$MFAB =MCAM +1$

Substitute

$MCAM =DCAB $

$MFAB =DCAB +1$

$MF=AB+DCAB⋅DC $

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.

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.

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

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

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

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

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.$DCAD =BCEB substitute DCAD =BCAB $

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

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.

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

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.$DCBD =ACBA $

SubstituteExpressions

Substitute expressions

$x13−x =125 $

Solve for x

MultEqn

LHS⋅12x=RHS⋅12x

12(13−x)=5x

Distr

Distribute 12

156−12x=5x

AddEqn

LHS+12x=RHS+12x

156=17x

DivEqn

$LHS/17=RHS/17$

$17156 =x$

RearrangeEqn

Rearrange equation

$x=17156 $

$DCBD =ECAE $

SubstituteExpressions

Substitute expressions

$17156 1765 =12−zz $

Solve for z

DivFracByFracD

$c/da/b =ba ⋅cd $

$1765 ⋅15617 =12−zz $

MultFrac

Multiply fractions

$15665 =12−zz $

MultEqn

LHS⋅156(12−z)=RHS⋅156(12−z)

65(12−z)=156z

Distr

Distribute 65

780−65z=156z

AddEqn

LHS+65z=RHS+65z

780=221z

DivEqn

$LHS/221=RHS/221$

$221780 =z$

ReduceFrac

$ba =b/13a/13 $

$1760 =z$

RearrangeEqn

Rearrange equation

$z=1760 $

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.

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

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

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.

It is given that BD divides the opposite side in proportion to the sides meeting at B.$DCAD =BCAB substitute DCAD =BCBE $

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

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

Show that CD bisects ∠ACB.

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

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.
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.
### Hint

### Solution

### Extra

Show that all segments are cut by the others into congruent parts.

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

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. ### Step 1

### Step 2

### Step 3

### Step 4

### Step 5

Using similar logic as in the previous steps, intersection points in similar positions can also be marked. ### 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.

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.

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.

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.

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.

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

Again, these bisect each other.

This shows that when considering every second segment in both directions, the segments cut each other into congruent pieces.

The solution above was based on finding midpoints again and again. Similar arguments can be used to prove the claim for 2, 4, 8, $16,…$, 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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