5. Triangles & Congruence

Lesson

Geometric constructions can be made by first drawing a set of circular arcs. These arcs constitute points of interest such as vertices, from which we can join to create a figure like an angle or a triangle.

Construction of isosceles triangle $ABC$ABC |

We want to be able to interpret how a construction was formed given a set of circular arcs. In the above image, the dotted arc comes from a circle centered at $A$`A` which passes two arbitrarily chosen points, $B$`B` and $C$`C`. The constructed triangle $\triangle ABC$△`A``B``C` is isosceles, since the line segments $\overline{AB}$`A``B` and $\overline{AC}$`A``C` are congruent radii of the same circle.

In a diagram, we assume that every arc belongs to a circle that is centered at a given or constructed point. In the following image, arc $1$1 comes from a circle centered at the given point $C$`C` while arcs $2$2 and $3$3 come from a circle centered at the constructed point $A$`A`.

Construction of arcs and vertices |

We can also identify the order that the arcs were drawn. In the image above, arc $1$1 must be created first to construct the point $A$`A`. Then the arcs $2$2 and $3$3 can be drawn second, since they center at $A$`A`.

We lastly assume that certain arcs have equal radii. For instance, the arcs that form $A$`A` and $B$`B` share the same radii. Also, the four arcs at $D$`D` and $E$`E` all share the same radii.

Identifying when arcs have equal radii is important, since it allows us to make geometric statements about our constructions. For instance, the line $\overleftrightarrow{AB}$›‹`A``B` bisects the line segment $\overline{DE}$`D``E`. We can show this using **congruent triangles**.

$\overleftrightarrow{AB}$›‹AB bisects $\overline{DE}$DE |

To prove that $\overleftrightarrow{AB}$›‹`A``B` bisects $\overline{DE}$`D``E`, we first want to show that the triangles $\triangle ADB$△`A``D``B` and $\triangle AEB$△`A``E``B` are congruent.

Note that $\overline{AD}$`A``D` is congruent to $\overline{AE}$`A``E` and $\overline{DB}$`D``B` is congruent to $\overline{EB}$`E``B` because the arcs that created these congruent segments have the same radii. Of course $\overline{AB}$`A``B` is a common side to both triangles, and so $\triangle ADB$△`A``D``B` is congruent to $\triangle AEB$△`A``E``B` by side-side-side congruence.

The next step is to show that $\triangle ADC$△`A``D``C` and $\triangle AEC$△`A``E``C` are congruent.

We already have a pair of sides that are congruent from before, $\overline{AD}$`A``D` and $\overline{AE}$`A``E`. $\overline{AC}$`A``C` is a common side to both triangles, and so is clearly congruent to itself. Lastly we know that $\angle DAC$∠`D``A``C` is congruent to $\angle EAC$∠`E``A``C` because they represent corresponding angles from the pair of congruent triangles, $\triangle ADC$△`A``D``C` and $\triangle AEC$△`A``E``C`. So $\triangle ADC$△`A``D``C` and $\triangle AEC$△`A``E``C` have side-angle-side congruence.

This means that $\overline{DC}$`D``C` is congruent to $\overline{EC}$`E``C` because they are corresponding sides of the congruent triangles, $\triangle ADC$△`A``D``C` and $\triangle AEC$△`A``E``C`. This means that $\overleftrightarrow{AB}$›‹`A``B` must bisect $\overline{DE}$`D``E`.

We can formalize the above explanation into a two column proof as shown below.

Statements |
Reasons |

$\overline{AD}\cong\overline{AE}\cong\overline{DB}\cong\overline{EB}$ |
The arcs that created the segments have the same radius. |

$\overline{AB}\cong\overline{AB}$AB≅AB |
Reflexive property of congruence |

$\triangle ADB\cong\triangle AEB$△ADB≅△AEB |
Side-side-side congruence of triangles |

$\overline{AC}\cong\overline{AC}$AC≅AC |
Reflexive property of congruence |

$\angle DAC\cong\angle EAC$∠DAC≅∠EAC |
Corresponding parts of congruent triangles are congruent (CPCTC). |

$\triangle ADC\cong\triangle AEC$△ADC≅△AEC |
Side-angle-side congruence of triangles |

$\overline{DC}\cong\overline{EC}$DC≅EC |
Corresponding parts of congruent triangles are congruent (CPCTC). |

$\overleftrightarrow{AB}$›‹AB bisects $\overline{DE}$DE |
Definition of bisector |

In the image below, $\angle EDF$∠`E``D``F` is constructed congruent to $\angle BAC$∠`B``A``C`. Justify the steps of construction in a two column proof.

Prove that the construction of $\overrightarrow{AD}$›‹`A``D` is an angle bisector of the given angle $\angle BAC$∠`B``A``C`.

Make formal geometric constructions with a variety of tools and methods. Constructions include: copying segments; copying angles; bisecting segments; bisecting angles; constructing perpendicular lines, including the perpendicular bisector of a line segment; and constructing a line parallel to a given line through a point not on the line.

Use congruence and similarity criteria to prove relationships in geometric figures and solve problems utilizing real-world context.