Triangle Proofs Workshop

Module 02 — Transformations and Congruence

The big idea Build the proof one justified step at a time. Then try it yourself.
Three evidence boxes feed SAS, which then supports a CPCTC conclusion.

By the end you can

  • Connect each statement to a valid reason.
  • Build and explain the six triangle-theorem proofs in your course.
  • Distinguish a theorem from its converse and use CPCTC in the correct order.

Start with the proof vocabulary

Use this if the statements make sense but the reasons feel unfamiliar.

Not sure about the basics underneath this lesson?

A proof is a chain of evidence

Batman, the goal is not to guess the teacher’s next sentence. Start with what you know, name what you must prove, and connect them with reasons.

Every statement needs a reason that actually supports it. Marks and givens count; “it looks equal” does not.

Triangle ABC with vertex A above base BC. AB and AC have matching tick marks. AD bisects angle BAC, meeting BC at D. The two angles at A have matching arcs.
Matching ticks are evidence. The apparent symmetry alone is not.

Read the destination first

Given: AB ≅ AC. Prove: ∠ABC ≅ ∠BCA.

Those are the angles at B and C, opposite the equal sides. Work backward: if you can prove two smaller triangles congruent, CPCTC will give you the matching angles.

Triangle ABC with vertex A above base BC. AB and AC have matching tick marks. AD bisects angle BAC, meeting BC at D. The two angles at A have matching arcs.
The middle letter names the angle’s vertex.

Three formats, the same logic

A two-column proof pairs each statement with its reason. A paragraph proof puts the same links into sentences. A flow proof uses arrows to show which facts support the next fact.

Changing the format does not remove any evidence. Try saying “because…” after every claim.

Three evidence boxes, AB congruent AC, angle BAD congruent CAD, and AD congruent AD, feed a SAS box; SAS feeds the base-angle conclusion by CPCTC.
A flow arrow means “this fact helps justify that conclusion.”

Triangle Sum: move the angles onto a line

Construct LR through A parallel to BC. AB and AC now act as transversals.

The angle at B matches ∠LAB; the angle at C matches ∠CAR, by alternate interior angles. These two angles and ∠BAC fill a straight angle at A.

Triangle ABC with A at the top, B bottom left, C bottom right. Line LA R passes through A parallel to BC. Angles LAB and ABC carry matching single arcs; angles CAR and BCA carry matching double arcs.
The matching arc marks track the same angles—not the same positions.

Build the Triangle Sum proof

Predict each reason, then reveal one row. We are proving Triangle Sum, so using Triangle Sum as a reason here would be circular.

Triangle ABC with A at the top, B bottom left, C bottom right. Line LA R passes through A parallel to BC. Angles LAB and ABC carry matching single arcs; angles CAR and BCA carry matching double arcs.
A construction + angle relationships + substitution proves the total.

Using a theorem is different from proving it

Once Triangle Sum is established, you may use it to find an unknown angle. With angles 27° and 100°, the third is 180° − 27° − 100° = 53°.

If a given 80° is an outside neighbor on a straight line, find the inside angle first: 180° − 80° = 100°. Do not subtract the outside 80° as though it were inside.

Triangle ABC with A at the top, B bottom left, C bottom right. Line LA R passes through A parallel to BC. Angles LAB and ABC carry matching single arcs; angles CAR and BCA carry matching double arcs.
Find the triangle’s three interior angles before adding.

Build the Isosceles Triangle proof

Given AB ≅ AC, construct AD as the bisector of ∠BAC. D lies on BC. We want the base angles at B and C to match.

Triangle ABC with vertex A above base BC. AB and AC have matching tick marks. AD bisects angle BAC, meeting BC at D. The two angles at A have matching arcs.
The shared side supplies the second S in SAS.
Not sure about this bit?

The theorem and its converse

Isosceles: AB ≅ AC → ∠B ≅ ∠C.

Converse: ∠B ≅ ∠C → AB ≅ AC.

Read what was given before choosing a direction. Equal sides lead to opposite equal angles; equal angles lead to opposite equal sides.

Triangle ABC with vertex A above base BC. AB and AC have matching tick marks. AD bisects angle BAC, meeting BC at D. The two angles at A have matching arcs.
Opposite AB is ∠C. Opposite AC is ∠B.

Prove the converse without assuming equal sides

Now suppose only ∠ABC ≅ ∠BCA. Construct the angle bisector AD. Do not mark AB and AC equal yet: that is what we must prove.

Triangle ABC with B and C marked as equal base angles. AD bisects angle BAC. No equal-side ticks appear on AB or AC.
No equal-side ticks until the proof earns them.

Two reasons, two jobs

If AB = AC and ∠A = 90°, Isosceles gives ∠B ≅ ∠C. Triangle Sum gives m∠B + m∠C = 90°.

Let each equal angle be x: 2x = 90°, so x = 45°. Neither theorem does both jobs alone.

Triangle ABC with vertex A above base BC. AB and AC have matching tick marks. AD bisects angle BAC, meeting BC at D. The two angles at A have matching arcs.
Equal sides explain equality. The angle total supplies the number.
Not sure about this bit?

Triangle Inequality: close the shape

For side lengths a, b, and c, each pair’s sum must be greater than the third: a + b > c, a + c > b, b + c > a.

For 5, 7, and 12, the two shorter sides add to exactly 12. The shape lies flat—it is not a triangle. With two sides 5 and 7, the third side x must satisfy 2 < x < 12.

Two diagrams: sides 5 and 7 laid along a straight segment of length 12, and a closed triangle with sides labeled 5, 7, and 10.
The endpoints 2 and 12 are excluded, not allowed.

Why two sides must beat the third

This workshop example adapts the course’s side–angle relationship. Extend AB beyond B to D so BD = BC. Then prove AB + BC > AC.

Triangle ABC with A and B along a horizontal line, C above. AB extends past B to D. BC and BD have matching ticks. Dashed CD makes isosceles triangle BCD.
Compare angles in ΔACD, not in two different triangles.

Midsegment: two midpoint facts

A midsegment joins the midpoints of two sides. It is parallel to the third side and half as long.

If M and N are the midpoints of AC and BC, then MN ∥ AB and MN = ½AB. If AB = 8, MN = 4. If MN = 4, work backward: AB = 8.

Triangle ABC at A(0,0), B(8,0), C(2,6). M(1,3) and N(5,3) are midpoints of AC and BC. MN is horizontal, parallel to AB.
Both endpoints must be midpoints.

A coordinate proof of Midsegment

Use the general triangle A(0,0), B(2b,0), C(2u,2v), where b > 0 and v > 0. The drawing uses one numeric example, but the algebra works for every such triangle.

Triangle ABC at A(0,0), B(8,0), C(2,6). M(1,3) and N(5,3) are midpoints of AC and BC. MN is horizontal, parallel to AB.
The numeric picture illustrates—not replaces—the general proof.

Median is not midsegment

A median joins a vertex to the midpoint of the opposite side. A midsegment joins two side midpoints.

The three medians meet at the centroid. To prove concurrency, draw two medians, call their intersection D, then show the line through the third vertex and D hits the remaining side at its midpoint.

Triangle WXY. A and B are midpoints of WY and XY. Medians XA and WB meet at D. Line YDP meets WX at C; P is outside the triangle with YD equal to DP. WP and PX complete parallelogram WDXP. C has no midpoint marks because that is the conclusion to prove.
The difficult step is proving WC = CX—not merely drawing YC.

Medians: prove the third one joins the meeting

Given A is the midpoint of WY and B is the midpoint of XY, XA and WB meet at D. Let line YD meet WX at C. Predict why each new construction helps.

Triangle WXY. A and B are midpoints of WY and XY. Medians XA and WB meet at D. Line YDP meets WX at C; P is outside the triangle with YD equal to DP. WP and PX complete parallelogram WDXP. C has no midpoint marks because that is the conclusion to prove.
This follows the course’s construction proof; the coordinate positions are illustrative.
Not sure about this bit?

Turn the proof into a paragraph

Since A and D are midpoints of YW and YP, AD is parallel to WP by Midsegment. Because A, D, and X are collinear, DX is parallel to WP. Similarly, DB—and therefore WD—is parallel to PX. Thus WDXP is a parallelogram. Its diagonals bisect each other, so WC = CX. C is the midpoint of WX, making YC a median through D. Therefore the three medians are concurrent.

Same evidence, now in sentences. Keep the reasons; do not replace them with “obviously.”

Triangle WXY. A and B are midpoints of WY and XY. Medians XA and WB meet at D. Line YDP meets WX at C; P is outside the triangle with YD equal to DP. WP and PX complete parallelogram WDXP. C has no midpoint marks because that is the conclusion to prove.
A paragraph can be short without skipping its essential links.

Catch the proof traps

CPCTC comes after congruence. It does not prove triangles congruent. ASA needs the side between its two angles; SAS needs the angle between its two sides.

Do not prove a theorem with itself. You may use established earlier results. Do not assume the conclusion, rely on a picture’s apparent size, or claim a midpoint merely because it looks centered.

Evidence from two sides and their included angle supports SAS; SAS then supports a matching-angle conclusion through CPCTC.
Evidence first → congruent triangles → matching parts.

Your proof lab

Choose the statement or reason that completes each link. Read the explanation after your answer; mistakes here are practice.

1 / 12 0 correct

Final checklist

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