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{{ printedBook.courseTrack.name }} {{ printedBook.name }} 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. $PQ PQ ∥SR=SR $ 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.

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. $△PQM≅△RSM $ Corresponding parts of congruent triangles are congruent. $PMQM ≅RM≅SM $ This completes the proof that $PR$ and $QS $ bisect each other.

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. $DCAB =DMMB =MCAM $

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. $EMAB =DMDB $ In this equality $DB$ can be replaced by $DM+MB.$ $EMAB =DMDM+MB =1+DMMB $ 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 $EM.$ This calculation gave an expression for the length of $EM$ in terms of the lengths of the bases of the trapezoid.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 $ 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.$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.

$DCAD =BCAB $

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. ${∠1≅∠2∠2≅∠4 ⇒∠1≅∠4 $ 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. ${∠1≅∠3∠1≅∠4 ⇒∠3≅∠4 $ This can be shown in the diagram.

Note that $△ACE$ is divided by $ℓ,$ which is parallel to $AE.$ Therefore, by the Triangle Proportionality Theorem, $ℓ$ divides the other two sides of this triangle proportionally. $DCAD =BCEB $ 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.

$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. $BC=5_{2}+12_{2} =13 $ 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 $BD=BC−DC.$ Furthermore, let $x$ represent $DC,$ as it is unknown. Then, since $BC$ was just found, it can be said that $BD=13−x.$

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. $DCBD =ACBA $ Substituting the expressions from the diagram gives an equation that can be solved for $x,$ which represents the length of $DC.$$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.

$DCAD =BCAB ⇒∠ABD≅∠CBD $

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 $ Because $BE$ is equal to $AB,$ by the Substitution Property of Equality $BE$ can be substituted for $AB$ in the proportion. $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. $∠AEB≅∠CBD∠EAB≅∠ABD $ Because $BE=AB,$ by the Isosceles Triangle Theorem $∠AEB$ is congruent to $∠EAB.$

Since $∠CBD$ and $∠EAB$ are both congruent to $∠AEB,$ by the Transitive Property of Congruence, it follows that $∠CBD$ and $∠EAB$ are congruent angles. ${∠AEB≅∠CBD∠AEB≅∠EAB ⇓∠CBD≅∠EAB $ By the same property, since $∠ABD$ and $∠CBD$ are both congruent to $∠EAB,$ they are congruent angles.

${∠CBD≅∠EAB∠EAB≅∠ABD ⇓∠ABD≅∠CBD $

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. $CBAC =64 =32 $ That equals the ratio of the two segments on the third side of the triangle. $DBAD =32 $ According to the converse of the Angle Bisector Theorem, this relationship between the proportions means that $CD$ bisects $∠ACB.$

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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