Back to Exercise: Explain triangle congruence criteria

Exercises: Triangle Congruence Criteria from Rigid Motions

Work through each section in order. For explanation problems, use complete sentences and reference rigid motions where relevant.

Grade 9·21 problems·~30 min·Common Core Math - HS Geometry·standard·hsg-co-b-8

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Warm-Up: Review What You Know

These problems review prerequisite skills from CO.B.6, CO.B.7, and Grade 8 geometry.

According to the definition from HSG.CO.B.7, two triangles are congruent if and only if:

All three pairs of corresponding angles are congruent.

There exists a sequence of rigid motions mapping one triangle onto the other.

All three pairs of corresponding sides are congruent.

The triangles have the same area.

CPCTC stands for "Corresponding Parts of Congruent Triangles are Congruent." In a geometric proof, CPCTC is used:

As a reason to establish that two triangles are congruent.

After congruence has been established, to conclude that specific pairs of sides or angles are equal.

To prove that a triangle is isosceles.

As a substitute for identifying which congruence criterion applies.

In triangle $\triangle PQR$ , $\angle P = 48^\circ$ and $\angle Q = 75^\circ$ . What is the measure of $\angle R$ ?

$57^\circ$

$123^\circ$

$67^\circ$

$48^\circ$

Fluency Practice

Two triangles have the following known congruent parts: $\overline{AB} \cong \overline{DE}$ , $\angle A \cong \angle D$ , and $\overline{AC} \cong \overline{DF}$ . The angle $\angle A$ is between sides $\overline{AB}$ and $\overline{AC}$ . Which criterion guarantees the triangles are congruent?

SSS

ASA

SAS

SSA — two sides and a non-included angle

In the SAS rigid-motion proof, after translating vertex $A$ to vertex $D$ and rotating so that $B$ maps to $E$ , what forces vertex $C''$ to land exactly on vertex $F$ ?

The included angle $\angle A \cong \angle D$ forces ray $DC''$ to point in the same direction as ray $DF$ , and the side length $AC = DF$ places $C''$ at the correct distance.

The triangle angle sum forces $\angle C'' = \angle F$ , which places $C''$ on $F$ .

Two circles centered at $D$ and $E$ intersect at $F$ , determining $C''$ .

A reflection over line $DE$ maps $C''$ to $F$ .

In the ASA proof, after aligning side $\overline{AB}$ onto $\overline{DE}$ (so $A'' = D$ and $B'' = E$ ), why is vertex $C''$ uniquely determined as $F$ ?

The side $AC = DF$ places $C''$ at the correct distance from $D$ .

The angle at $A$ forces ray $DC''$ onto ray $DF$ , and the angle at $B$ forces ray $EC''$ onto ray $EF$ ; their intersection is uniquely $F$ .

Two circles centered at $D$ and $E$ with radii $DF$ and $EF$ intersect at $F$ .

The third angle $\angle C = \angle F$ is automatically equal, forcing the vertex to coincide.

In $\triangle ABC$ and $\triangle DEF$ , you know $\angle A \cong \angle D$ , $\angle C \cong \angle F$ , and $\overline{BC} \cong \overline{EF}$ . This is AAS (two angles and a non-included side). Which statement correctly explains why these triangles must be congruent?

AAS is a separate valid criterion, proved independently of ASA.

AAS does not guarantee congruence because the side is not included between the two angles.

Since $\angle A + \angle C + \angle B = 180^\circ$ and $\angle D + \angle F + \angle E = 180^\circ$ , we get $\angle B \cong \angle E$ ; now $\angle B \cong \angle E$ , $\overline{BC} \cong \overline{EF}$ , and $\angle C \cong \angle F$ is ASA applied to side $\overline{BC}$ .

AAS fails like SSA because neither has the side between the two known angles.

$Two overlapping circles centered at D and E. Their two intersection points, F and F-prime, are symmetric about line DE, illustrating the two possible positions for vertex C-double-prime in the SSS proof.$

In the SSS rigid-motion proof for $\triangle ABC \cong \triangle DEF$ (with $AB = DE$ , $AC = DF$ , $BC = EF$ ), after translating $A$ to $D$ and rotating so $B$ maps to $E$ , vertex $C''$ must satisfy $DC'' = DF$ and $EC'' = EF$ . Why does this guarantee that $C''$ is either $F$ or the reflection of $F$ over line $DE$ ?

Because the angle sum forces $\angle DC''E = \angle DFE$ .

Because $C''$ lies on two circles — one centered at $D$ with radius $DF$ , one centered at $E$ with radius $EF$ — and two distinct circles intersect in at most two points, which are symmetric about the line through their centers.

Because SAS applied to the two sub-triangles forces $C'' = F$ .

Because a translation followed by a rotation always produces a unique image.

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