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, justify, and apply formal geometric constructions.

Use congruence and similarity criteria for triangles to solve problems algebraically and geometrically.

Use congruence and similarity criteria for triangles to prove relationships in geometric figures.