Task Sets: Explain triangle congruence criteria - Common Core Math - HS Geometry

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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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Exercises: Triangle Congruence Criteria from Rigid Motions

Warm-Up: Review What You Know

Fluency Practice