How to Write a Geometry Proof (Step-by-Step Method)
Geometry proofs are where many students hit a wall. They feel different from algebra because you can’t just “do the math.” The fix is a consistent 5-step approach.
Quick answer
(1) Identify what’s given. (2) Identify what you need to prove. (3) Mark up the figure. (4) Plan the path from given to goal. (5) Write the two-column proof — each step justified.
Step 1 — Identify what’s given
The problem will tell you. Highlight every “given” statement. These are your starting facts.
Step 2 — Identify what you need to prove
The conclusion you’re working toward. Often it’s an equality, a congruence, or parallel/perpendicular relationship.
Step 3 — Mark up the figure
Draw congruent sides with tick marks. Congruent angles with arcs. Parallel lines with arrows. Right angles with squares. This visual gives your brain something to work with.
Step 4 — Plan the path
Work backwards from the goal. “I need to prove these triangles are congruent. To do that I need SSS, SAS, ASA, AAS, or HL. Which do I have?” Then forward from given: “I have these two pairs of sides congruent and this angle. That’s SAS.”
Step 5 — Write the two-column proof
Left column: statements. Right column: reasons. Every step must be justified by either a given, a definition, a postulate, or a theorem.
Common reasons (memorize)
- Reflexive property: a segment/angle equals itself
- Vertical angles theorem: vertical angles are congruent
- Triangle sum: angles of a triangle add to 180°
- SSS, SAS, ASA, AAS, HL: triangle congruence
- CPCTC: corresponding parts of congruent triangles are congruent
- Definition of [thing]: e.g. definition of midpoint, definition of perpendicular
Worked example
Given: Triangle ABC with AB = AC. Point M is the midpoint of BC.
Prove: AM ⊥ BC.
The plan
Show triangles ABM and ACM are congruent (SSS). Then angles AMB and AMC are congruent (CPCTC). Since they form a straight line, they’re each 90°. So AM ⊥ BC.
Two-column proof
| Statement | Reason |
|---|---|
| AB = AC | Given |
| M is midpoint of BC | Given |
| BM = MC | Definition of midpoint |
| AM = AM | Reflexive property |
| △ABM ≅ △ACM | SSS |
| ∠AMB ≅ ∠AMC | CPCTC |
| ∠AMB + ∠AMC = 180° | Linear pair |
| 2(∠AMB) = 180°, ∠AMB = 90° | Substitution and division |
| AM ⊥ BC | Definition of perpendicular |
Common mistakes
- Forgetting reflexive property when a side or angle is shared.
- Using SSA (not valid for proving congruence).
- Not labeling congruent parts on the figure.
- Reasons that are too vague (“because” instead of “vertical angles theorem”).