Honors Geometry Unit 4 Review: Congruent Triangles

A congruence proof is a chain of statements where every step has a reason. Reasons come from the givens, definitions, postulates, and theorems you've already proven. Most proofs in this unit follow the same arc. You prove two triangles congruent, then use CPCTC to get the side or angle the problem actually asked about.

Start by marking the diagram with tick marks for congruent sides and arcs for congruent angles. Then count your pairs, name the criterion, and write the congruence statement with vertices in matching order.

Worked example. Given $\overline{AB} \cong \overline{CB}$ and $\overline{BD}$ bisects $\angle ABC$, with $D$ on $\overline{AC}$. Prove $\overline{AD} \cong \overline{CD}$.

1. $\overline{AB} \cong \overline{CB}$ (Given)
2. $\angle ABD \cong \angle CBD$ (Definition of angle bisector)
3. $\overline{BD} \cong \overline{BD}$ (Reflexive Property)
4. $\triangle ABD \cong \triangle CBD$ (SAS)
5. $\overline{AD} \cong \overline{CD}$ (CPCTC)

The angle in step 2 sits between the sides in steps 1 and 3, so SAS applies. A paragraph proof uses the same logic written as connected sentences, such as "Since $\overline{BD}$ bisects $\angle ABC$, we know $\angle ABD \cong \angle CBD$."

Triangles in a diagram can overlap or share a side or angle, which can make them hard to see. Redraw the two triangles separately, keeping the vertex labels the same. Any side or angle they share is congruent to itself by the Reflexive Property, so it may provide a pair you were missing.

An equilateral triangle has three congruent sides and three $60^\circ$ angles. Its altitude to any side is also a median, an angle bisector, and the perpendicular bisector of that side. The altitude splits it into two 30-60-90 triangles, so a side of length $s$ gives an altitude of $\frac{s\sqrt{3}}{2}$.

Equilateral triangles are a special case of isosceles triangles. The Isosceles Triangle Theorem says the angles opposite congruent sides are congruent, and its converse is also true.

Worked example. An equilateral triangle has sides $2x + 1$, $3x - 4$, and $x + 6$. Find $x$, the side length, and the altitude.

Set two sides equal, $2x + 1 = 3x - 4$, which gives $x = 5$. Check all three sides with $x = 5$. You get $2(5) + 1 = 11$, $3(5) - 4 = 11$, and $5 + 6 = 11$. Each side is 11, so the altitude is $\frac{11\sqrt{3}}{2} \approx 9.53$.

Two equilateral triangles are always similar, but they are congruent only when their sides match.

A right triangle already gives you one known angle, $90^\circ$, so you need less extra information. The Hypotenuse-Leg (HL) Theorem says that if the hypotenuse and one leg of a right triangle are congruent to the hypotenuse and a leg of another right triangle, the triangles are congruent.

HL looks like SSA, which normally fails. It works here because the right angle allows only one triangle. With the hypotenuse and one leg fixed, the Pythagorean theorem forces the third side to match too. You must show both triangles are right triangles before citing HL.

Other right-triangle shortcuts are renamed versions of criteria you already know. Leg-Leg is SAS with the right angle included.

Worked example. $\triangle PQR$ is right at $Q$ with $PR = 13$ and $PQ = 5$. $\triangle XYZ$ is right at $Y$ with $XZ = 13$ and $YZ = 12$. Are they congruent?

Find the missing leg of $\triangle PQR$ with $a^2 + b^2 = c^2$. You get $5^2 + QR^2 = 13^2$, so $QR^2 = 169 - 25 = 144$ and $QR = 12$. Now the hypotenuses match ($PR = XZ = 13$) and a pair of legs match ($QR = YZ = 12$). By HL, $\triangle PQR \cong \triangle XYZ$. Write the vertices so the right angles ($Q$ and $Y$) and the matching legs line up.

SSS, SAS, and ASA are postulates, so they are accepted without proof. AAS and HL are theorems, because they can be proven from earlier results. AAS follows from ASA. If two angles of one triangle match two angles of another, the third angles must match too, since every triangle's angles sum to $180^\circ$. That gives you ASA.

The word "included" decides which criterion fits. In $\triangle ABC$, the angle included between $\overline{AB}$ and $\overline{BC}$ is $\angle B$, and the side included between $\angle A$ and $\angle B$ is $\overline{AB}$.

SSA and AAA do not prove congruence. AAA fixes the shape but not the size, so a triangle with sides 3, 4, 5 and one with sides 6, 8, 10 have the same angles but are not congruent. SSA can produce two different triangles, called the ambiguous case, because the third side can swing into two positions.

Congruence can also be defined with rigid motions. Two triangles are congruent if a sequence of translations, rotations, and reflections maps one exactly onto the other. Rigid motions preserve distances and angle measures, which is why every criterion works.