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Geometry Proof Solver

Two-column proofs with every statement paired to the theorem that justifies it
Given that M is the midpoint of AB and of CD, prove that triangle AMC ≅ triangle BMD
Prove that vertical angles are congruent
Prove that the exterior angle of a triangle equals the sum of the two remote interior angles
Given AB ≅ AC, prove that angle B ≅ angle C

How a Two-Column Proof Works

A proof is a chain in which every statement is licensed by something already true: the given, a definition, a postulate, a previously proved theorem, or an algebraic property. The two-column layout makes the licence explicit — statements on the left, reasons on the right.

The method:

  1. Mark the diagram with everything given. Congruent marks, right-angle boxes and shared sides are what make the next step visible.
  2. Work backwards from the goal. To prove two segments congruent, the usual route is to prove the triangles containing them congruent, then finish with CPCTC.
  3. Collect the ingredients the chosen theorem needs — SSS, SAS, ASA, AAS or HL each demand three specific parts in a specific arrangement.
  4. Include the free facts: a shared side is congruent to itself (Reflexive Property), and intersecting lines create congruent vertical angles.
  5. Write the reason for every line. A statement with no reason is a gap, not a shortcut.

The Reasons You Will Actually Use

ReasonWhat it lets you write
Definition of midpointThe midpoint splits a segment into two congruent halves
Definition of bisectorAn angle bisector makes two congruent angles
Reflexive PropertyAB‾≅AB‾\overline{AB} \cong \overline{AB} — a shared side or angle
Vertical Angles TheoremOpposite angles at an intersection are congruent
Linear Pair / SupplementaryTwo adjacent angles on a line sum to 180∘180^\circ
Alternate Interior AnglesRequires parallel lines cut by a transversal
Triangle Sum TheoremThe three angles of a triangle sum to 180∘180^\circ
SSS, SAS, ASA, AAS, HLTriangle congruence
CPCTCOnce triangles are congruent, any corresponding parts are

The condition people forget: SAS needs the angle between the two sides, and ASA needs the side between the two angles. Get the arrangement wrong and you have SSA, which does not prove congruence. HL applies only to right triangles, and CPCTC may only be used after the congruence is established — never as a step towards it.

Common Mistakes to Avoid

  • Assuming from the picture. Segments that look equal, angles that look right and lines that look parallel prove nothing unless marked or given.
  • Using SSA or AAA. Neither establishes congruence; AAA gives similarity only.
  • Using CPCTC too early. It is the conclusion drawn from congruent triangles, not a reason for congruence.
  • Applying parallel-line theorems without parallel lines. Alternate interior and corresponding angles need the parallel condition stated or proved first.
  • Misordering the vertices. △ABC≅△DEF\triangle ABC \cong \triangle DEF asserts A↔DA \leftrightarrow D, B↔EB \leftrightarrow E, C↔FC \leftrightarrow F; a scrambled order makes CPCTC produce false claims.
  • Leaving a reason blank or writing "obvious". Every line needs a named definition, postulate or theorem.

示例题目

Step 1: ∠APC\angle APC and ∠APD\angle APD form a linear pair, so m∠APC+m∠APD=180∘m\angle APC + m\angle APD = 180^\circ — Linear Pair Postulate
Step 2: ∠APD\angle APD and ∠BPD\angle BPD form a linear pair, so m∠APD+m∠BPD=180∘m\angle APD + m\angle BPD = 180^\circ — Linear Pair Postulate
Step 3: Therefore m∠APC+m∠APD=m∠APD+m∠BPDm\angle APC + m\angle APD = m\angle APD + m\angle BPD — Substitution
Step 4: Subtract m∠APDm\angle APD from both sides: m∠APC=m∠BPDm\angle APC = m\angle BPD — Subtraction Property of Equality
Step 5: Equal measures mean congruent angles — Definition of congruent angles
Answer: ∠APC≅∠BPD\angle APC \cong \angle BPD, proved

Step 1: AM‾≅MB‾\overline{AM} \cong \overline{MB} — Definition of midpoint (given, MM bisects AB‾\overline{AB})
Step 2: CM‾≅MD‾\overline{CM} \cong \overline{MD} — Definition of midpoint (given, MM bisects CD‾\overline{CD})
Step 3: ∠AMC≅∠BMD\angle AMC \cong \angle BMD — Vertical Angles Theorem, since ABAB and CDCD meet at MM
Step 4: The congruent angle lies between the two pairs of congruent sides, so the arrangement is side-angle-side
Step 5: △AMC≅△BMD\triangle AMC \cong \triangle BMD — SAS Congruence Postulate
Step 6: AC‾≅BD‾\overline{AC} \cong \overline{BD} — CPCTC
Answer: △AMC≅△BMD\triangle AMC \cong \triangle BMD by SAS, and AC‾≅BD‾\overline{AC} \cong \overline{BD} by CPCTC

Step 1: m∠A+m∠B+m∠ACB=180∘m\angle A + m\angle B + m\angle ACB = 180^\circ — Triangle Sum Theorem
Step 2: ∠ACB\angle ACB and ∠ACD\angle ACD form a linear pair on line BDBD, so m∠ACB+m∠ACD=180∘m\angle ACB + m\angle ACD = 180^\circ — Linear Pair Postulate
Step 3: Set the two sums equal: m∠A+m∠B+m∠ACB=m∠ACB+m∠ACDm\angle A + m\angle B + m\angle ACB = m\angle ACB + m\angle ACD — Substitution
Step 4: Subtract m∠ACBm\angle ACB from both sides — Subtraction Property of Equality
Step 5: Numerical check: if m∠A=55∘m\angle A = 55^\circ and m∠B=70∘m\angle B = 70^\circ, then m∠ACB=55∘m\angle ACB = 55^\circ and m∠ACD=180∘−55∘=125∘=55∘+70∘m\angle ACD = 180^\circ - 55^\circ = 125^\circ = 55^\circ + 70^\circ
Answer: m∠ACD=m∠A+m∠Bm\angle ACD = m\angle A + m\angle B, the Exterior Angle Theorem, proved

常见问题

A two-column proof lists each statement on the left and the reason that justifies it on the right. The reasons may only be the given information, definitions, postulates, previously proved theorems, or properties of equality — every line needs one.

CPCTC — corresponding parts of congruent triangles are congruent — is used only after you have already proved the triangles congruent by SSS, SAS, ASA, AAS or HL. It is the last step of a proof, never a reason for the congruence itself.

Two sides and a non-included angle can describe two genuinely different triangles — the ambiguous case, the same one that makes the law of sines produce two answers. Only SSS, SAS, ASA, AAS and (for right triangles) HL guarantee congruence.

No. A diagram may only be used for what is marked or stated: congruence marks, right-angle boxes, parallel arrows, and betweenness of points on a line. Equal-looking lengths, right-looking angles and parallel-looking lines must be given or proved.

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