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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.$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 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.

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 =17156 =13−17156 =1765 $

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 $EC=12−z.$
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.
$DCBD =ECAE $

Substituting the expressions from the diagram gives an equation that can be solved for $z,$ which represents the length of $AE.$
$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.

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 $∠ACB.$