Math Core

So far, proving two triangles congruent has been the goal. Usually, though, it's a stepping stone. Once you know two triangles are congruent, every pair of corresponding parts is congruent, including the parts you weren't told anything about. That idea lets you prove that segments are equal, angles are equal, and even measure distances you can't reach.

What CPCTC means

Remember the definition of congruent triangles from the start of this unit: all six pairs of corresponding parts are congruent. The shortcuts (SSS, SAS, ASA, AAS, HL) let you establish congruence from only three pairs. After that, the definition hands you the other three pairs for free.

CPCTC

Corresponding Parts of Congruent Triangles are Congruent.

Once you have proved $\triangle ABC \cong \triangle DEF$ , you may conclude any of $\overline{AB} \cong \overline{DE}$ , $\overline{BC} \cong \overline{EF}$ , $\overline{AC} \cong \overline{DF}$ , $\angle A \cong \angle D$ , $\angle B \cong \angle E$ , $\angle C \cong \angle F$ , with the reason "CPCTC."

A three-step strategy

When a proof asks you to show two segments or two angles are congruent, try this plan.

Find two triangles that contain the parts you want, one part in each triangle, in corresponding positions.
Prove those triangles congruent with a shortcut, using the givens and any hidden facts (shared sides, vertical angles, parallel lines).
Finish with CPCTC.

Common mistake

CPCTC can only be used after a line in your proof that says the triangles are congruent. A common mistake is to use CPCTC to get one of the three parts you need for SSS, SAS or another shortcut. That's circular: you would be using the conclusion to prove itself. Also, CPCTC is not a shortcut itself, so it never proves triangles congruent.

Worked example: Proving segments congruent

Given: $M$ is the midpoint of $\overline{AE}$ and of $\overline{BD}$ .

Prove: $\overline{AB} \cong \overline{ED}$

Segments AE and BD cross at M, the midpoint of both.

$\overline{AB}$ is in $\triangle AMB$ and $\overline{ED}$ is in $\triangle EMD$ , so prove those triangles congruent first.

Statement	Reason
1. $M$ is the midpoint of $\overline{AE}$ and of $\overline{BD}$	Given
2. $\overline{AM} \cong \overline{EM}$ and $\overline{BM} \cong \overline{DM}$	Definition of midpoint
3. $\angle AMB \cong \angle EMD$	Vertical angles are congruent
4. $\triangle AMB \cong \triangle EMD$	SAS
5. $\overline{AB} \cong \overline{ED}$	CPCTC

In the same way you could also conclude $\angle A \cong \angle E$ or $\angle B \cong \angle D$ .

Worked example: Proving a pair of sides congruent in a quadrilateral

Given: $\overline{AB} \parallel \overline{DC}$ and $\overline{AB} \cong \overline{DC}$

Prove: $\overline{BC} \cong \overline{DA}$

AB is parallel to DC, AB = DC, and AC is a diagonal.

The diagonal $\overline{AC}$ splits the quadrilateral into $\triangle BAC$ and $\triangle DCA$ . $\overline{BC}$ is in the first and $\overline{DA}$ is in the second.

Statement	Reason
1. $\overline{AB} \parallel \overline{DC}$	Given
2. $\angle BAC \cong \angle DCA$	Alternate interior angles theorem
3. $\overline{AB} \cong \overline{DC}$	Given
4. $\overline{AC} \cong \overline{CA}$	Reflexive property of congruence
5. $\triangle BAC \cong \triangle DCA$	SAS
6. $\overline{BC} \cong \overline{DA}$	CPCTC

In step 5, the angle at $A$ in $\triangle BAC$ is between $\overline{BA}$ and $\overline{AC}$ , and the angle at $C$ in $\triangle DCA$ is between $\overline{DC}$ and $\overline{CA}$ . So it's the included angle, and SAS applies. You'll see this result again in the Quadrilaterals unit.

CPCTC with measurements

When congruent triangles have measurements attached, CPCTC turns into an equation.

Worked example: Measuring across a pond

A surveyor wants the distance $AB$ across a pond. She picks a point $C$ on dry land, walks from $A$ through $C$ to a point $D$ with $CD = AC$ , and from $B$ through $C$ to a point $E$ with $CE = BC$ . Then she measures $DE$ .

C is the midpoint of AD and of BE. Measuring DE gives AB.

Why it works. $\overline{AC} \cong \overline{DC}$ and $\overline{BC} \cong \overline{EC}$ by construction, and $\angle ACB \cong \angle DCE$ because they are vertical angles. So $\triangle ACB \cong \triangle DCE$ by SAS, and $\overline{AB} \cong \overline{DE}$ by CPCTC.

With algebra. Suppose a diagram of this setup labels $DE = 2x + 18$ and $AB = 5x - 3$ (in meters). Since $AB = DE$ ,

5x - 3 = 2x + 18 \quad\Rightarrow\quad 3x = 21 \quad\Rightarrow\quad x = 7,

so $AB = 5(7) - 3 = 32$ meters. (Check: $DE = 2(7) + 18 = 32$ .)

Tip

When a proof asks for congruent segments or angles, don't start with the parts. Start by circling the two triangles that hold them. The rest of the proof is just proving those triangles congruent.

Practice

Practice 1

Given $\triangle RST \cong \triangle UVW$ , which conclusion follows by CPCTC?

Practice 2

In the proof that $\overline{AB} \cong \overline{ED}$ (where $M$ is the midpoint of $\overline{AE}$ and $\overline{BD}$ ), what is the reason for the last statement, $\overline{AB} \cong \overline{ED}$ ?

Practice 3

Jordan knows that $\overline{AB} \cong \overline{DE}$ , $\overline{BC} \cong \overline{EF}$ and $\angle A \cong \angle D$ . He concludes $\angle C \cong \angle F$ by CPCTC. What is wrong with his reasoning?

Practice 4

Given: $\overline{AC}$ bisects $\angle BAD$ and $\angle BCD$ . Prove: $\overline{AB} \cong \overline{AD}$ .

AC bisects angle BAD and angle BCD.

Statement	Reason
1. $\overline{AC}$ bisects $\angle BAD$ and $\angle BCD$	Given
2. $\angle BAC \cong \angle DAC$	Definition of angle bisector
3. $\overline{AC} \cong \overline{AC}$	Reflexive property of congruence
4. $\angle BCA \cong \angle DCA$	Definition of angle bisector
5. $\triangle BAC \cong \triangle DAC$	?
6. $\overline{AB} \cong \overline{AD}$	CPCTC

What is the missing reason in step 5?

Practice 5

In the figure, $\overline{AB} \parallel \overline{DC}$ and $\overline{AB} \cong \overline{DC}$ , so $\triangle BAC \cong \triangle DCA$ . If $AD = 2y + 3$ and $BC = 4y - 9$ , find $AD$ .

AB is parallel to DC, AB = DC, and AC is a diagonal.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 6

In the pond setup, $C$ is the midpoint of $\overline{AD}$ and of $\overline{BE}$ . If $AB = 5x - 19$ and $DE = 3x + 5$ , find $AB$ .

C is the midpoint of AD and of BE. Measuring DE gives AB.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 7

In quadrilateral $ABCD$ , $\overline{AB} \cong \overline{CB}$ and $\overline{AD} \cong \overline{CD}$ , with diagonal $\overline{BD}$ . If $m\angle ABD = 35^\circ$ and $m\angle ADB = 40^\circ$ , find $m\angle BCD$ in degrees. (The figure is not drawn to scale.)

Quadrilateral ABCD with diagonal BD.

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 8

You want to prove $\angle P \cong \angle S$ , where $\angle P$ is in $\triangle PQR$ and $\angle S$ is in $\triangle STU$ . Which must appear in your proof before the line " $\angle P \cong \angle S$ , CPCTC"?