CPCTC & First Proofs

Once you know two triangles are congruent, you get a whole list of equal parts for free. That is what stands for: Corresponding Parts of Congruent Triangles are Congruent. If one triangle can be carried exactly onto the other, then each side lands on a matching side and each angle lands on a matching angle — so all six pairs are equal.

The one new skill is reading the correspondence. A congruence statement lists its vertices in matching order on purpose:

$\triangle ABC \cong \triangle DEF \quad\Longrightarrow\quad A\leftrightarrow D,\ \ B\leftrightarrow E,\ \ C\leftrightarrow F.$

From that single line you can read off every equal part — pair the letters in the same position, never the parts that happen to look closest on the page.

Read the correspondence

The second triangle below is drawn turned and flipped, so you cannot trust position. Click a side or angle of the first triangle and watch its true partner light up in the second — matched by the order $A\!\to\!D,\ B\!\to\!E,\ C\!\to\!F$ .

Given △ABC ≅ △DEF. The second triangle is turned and flipped — click a side or angle of the first triangle and read off its corresponding part by the letters, not the position.

Side AB of △ABC corresponds to side DE of △DEF. Both are marked with 1 tick in the same colour, so they read as corresponding even though the second triangle is reoriented.

Two congruent triangles under the stated correspondence △ABC ≅ △DEF, with the second drawn rotated and reflected. Selecting side AB highlights its corresponding part DE in the second triangle — matched by the correspondence order, not by where it sits.

Make a prediction

Make a prediction: Given △ABC ≅ △DEF, which side of the second triangle is congruent to BC?

DEEFDFIt depends on how the triangle is drawn

Why the order matters

CPCTC does not ask you to re-prove anything — the triangles are already congruent. It asks you to name the right partner. The trap is matching by appearance: a reflected triangle can put $E$ where you expect $D$ , so a part that looks like it pairs with $AB$ may really be $EF$ . Reading straight from $\triangle ABC \cong \triangle DEF$ keeps you right every time:

Sides: $AB \cong DE$ , $\;BC \cong EF$ , $\;CA \cong FD$ .
Angles: $\angle A \cong \angle D$ , $\;\angle B \cong \angle E$ , $\;\angle C \cong \angle F$ .

Key result

CPCTC: if $\triangle ABC \cong \triangle DEF$ , then every corresponding pair is congruent. Find a partner by pairing the letters in the same position of the congruence statement — not by which parts sit closest on the diagram.

Worked example: Given △PQR ≅ △STU, find the partner of ∠R and of side PQ.

Line the statements up by position: $P\leftrightarrow S$ , $Q\leftrightarrow T$ , $R\leftrightarrow U$ . The third letters match, so $\angle R \cong \angle U$ .
Before the next step: Which angle is congruent to ∠R?
∠S∠T∠U

Using CPCTC in a proof

CPCTC is almost always the last line of a proof. The strategy is a fixed two-step move:

Get a triangle congruence. Find two triangles that contain the parts you care about, and prove them congruent by one of the criteria — SSS, SAS, ASA, AAS, or HL.
Harvest a part with CPCTC. Once the triangles are congruent, the specific side or angle you actually wanted is congruent because corresponding parts of congruent triangles are congruent.

So before you can write CPCTC, you have to choose the right criterion from the given parts. That single decision — which three parts are given, and what do they spell? — is the whole game.

Is it congruent? — the checker

Practice the criterion-choosing move on five marked pairs. Four are genuine congruences; one is a deliberate trap. The verdict comes straight from the figure's coordinates, so it cannot drift from the math.

Each pair shows two triangles with their given parts marked (matching colour and tick/arc count = a given pair of equal parts). Pick a pair and decide: do those givens force the triangles to be congruent — and by which criterion?

Congruent — by SAS, △ABC ≅ △DEF

Two sides and the angle between them match — SAS forces the third side, so they are congruent. The verdict is computed from the triangles' coordinates, and the criterion SAS is read from the marked given parts.

Two triangles with their given parts marked. The checker reports whether they are congruent — sourced from the figure's coordinates — and, when they are, the criterion those givens certify (△ABC ≅ △DEF in corresponding order).

Make a prediction: One pair marks two sides and an angle that is NOT between them (SSA), yet the two triangles look different. Are they congruent?

Yes — two sides and an angle is always enoughNo — SSA does not force congruenceOnly if the angle is obtuse

Three worked proofs

Each proof makes the same two decisions — which criterion? then what does CPCTC give? Predict each before you reveal the finished two-column proof.

Proof 1 — SSS, then CPCTC

Given: $AB \cong CB$ , and $M$ is the midpoint of $\overline{AC}$ . Prove: $\angle A \cong \angle C$ .

The two triangles to compare are $\triangle ABM$ and $\triangle CBM$ — they share the segment $BM$ , and the midpoint splits $AC$ into two equal halves.

Make a prediction: Which criterion proves △ABM ≅ △CBM?

SSSSASASAAAS

Make a prediction: Now that △ABM ≅ △CBM, what does CPCTC let you conclude?

∠A ≅ ∠C∠A ≅ ∠BAM ≅ BMNothing more

Now build it: each statement is given in order — supply the reason that justifies it.

Given: $\overline{AB} \cong \overline{CB}$ , and $M$ is the midpoint of $\overline{AC}$

Prove: $\angle A \cong \angle C$

Statements	Reasons
1.
2.
3.
4.
5.
6.

Row 1: choose the reason that justifies this statement.

Reasons

Placed 0 of 6 steps.

Build a two-column proof that

\angle A \cong \angle C

: each statement is given in order and you supply the reason that justifies it. Each row is checked against the logic of the proof, so only a step that genuinely follows will seat.

Proof 2 — SAS, then CPCTC

Given: $\overline{AC}$ and $\overline{BD}$ bisect each other at $E$ . Prove: $AB \cong CD$ .

"Bisect each other" means $E$ is the midpoint of both diagonals, so it splits each into two equal pieces. Compare $\triangle AEB$ and $\triangle CED$ .

Make a prediction: Which criterion proves △AEB ≅ △CED?

SSSSASASAHL

Make a prediction: Once △AEB ≅ △CED, what does CPCTC give you?

AB ≅ CDAE ≅ BE∠B ≅ ∠CNothing more

Given: $\overline{AC}$ and $\overline{BD}$ bisect each other at $E$

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

Statements	Reasons
1.
2.
3.
4.
5.
6.

Row 1: choose the reason that justifies this statement.

Reasons

Placed 0 of 6 steps.

Build a two-column proof that

\overline{AB} \cong \overline{CD}

: each statement is given in order and you supply the reason that justifies it. Each row is checked against the logic of the proof, so only a step that genuinely follows will seat.

Proof 3 — ASA, then CPCTC

Given: $AB \parallel CD$ and $AD \parallel BC$ . Prove: $AB \cong CD$ .

Two pairs of parallel sides make $ABCD$ a parallelogram. Draw the diagonal $\overline{AC}$ and compare $\triangle ABC$ with $\triangle CDA$ .

Make a prediction: Which criterion proves △ABC ≅ △CDA?

SSSSASASASSA

Make a prediction: With △ABC ≅ △CDA established, what does CPCTC conclude?

AB ≅ CDAC ≅ BD∠B ≅ ∠ANothing more

Given: $\overline{AB} \parallel \overline{CD}$ , and $\overline{AD} \parallel \overline{BC}$

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

Statements	Reasons
1.
2.
3.
4.
5.
6.
7.

Row 1: choose the reason that justifies this statement.

Reasons

Placed 0 of 7 steps.

Build a two-column proof that

\overline{AB} \cong \overline{CD}

: each statement is given in order and you supply the reason that justifies it. Each row is checked against the logic of the proof, so only a step that genuinely follows will seat.

Key result

To prove two parts congruent, find two triangles that contain them, prove the triangles congruent by a criterion, then finish with CPCTC. The criterion you pick is dictated by the given parts: three sides → SSS, two sides and the included angle → SAS, two angles and the included side → ASA. These three proofs are the classic results behind isosceles base angles, bisected diagonals, and the opposite sides of a parallelogram.

For teachers

Standard, common misconceptions, and suggested use will live here.