MathIsimple
Lesson 3.3: Proving Triangle Congruence

From Given to Proven

Use definitions, theorems, and constructions to write rigorous proofs. Learn two-column, flow, and paragraph proof styles.

Learning Objectives

  • Organize proofs logically
  • Use auxiliary lines (altitudes, medians)
  • Apply congruence criteria in proofs
  • Conclude perpendicular/parallel results

Proof Strategies

Auxiliary Lines Catalog

  • Altitude: create right angles (⊥\perp)
  • Median: create midpoint relations
  • Angle bisector: split an angle equally
  • Perpendicular bisector: equal distances to endpoints

Flow of Reasoning

  • Translate givens into equalities (segments/angles)
  • Choose criteria (SAS/ASA/SSS/AAS/HL)
  • Conclude congruence → transfer equal parts (CPCTC)
  • Derive target (e.g., ⊥\perp, ∥\parallel, midpoint)

Proof Example

In △ABC with AB=ACAB=AC and AD the median, prove AD⊥BCAD\perp BC.

Given: AB=ACAB=AC, AD is a median (BD=DCBD=DC). Prove: AD⊥BCAD\perp BC.

Consider △ABD and △ACD: AB=ACAB=AC, BD=DCBD=DC, AD=ADAD=AD → SSS → △ABD ≅ △ACD → ∠ADB=∠ADC\angle ADB=\angle ADC.

Since ∠ADB\angle ADB and ∠ADC\angle ADC are linear pair and equal, each is 90°. Hence AD⊥BCAD\perp BC.

Alternative Styles

Paragraph Proof (Sketch)

Translate A to A', use midpoint property of median, apply SSS to △ABD and △ACD, deduce equal base angles, hence AD ⟂ BC.

Flow Proof (Outline)

Givens → equal segments → SSS → triangle congruence → CPCTC → right angles → perpendicularity.

Common Pitfalls

  • Using CPCTC before establishing triangle congruence.
  • Assuming medians/altitudes/angle bisectors are interchangeable without proof.
  • Forgetting linear pair/supplementary angle facts when concluding 90°.

Practice

1) In △ABC, AB=AC and E is midpoint of BC. Prove AE ⟂ BC.

Hint
Use SSS on △ABE and △ACE, then linear pair.

2) Construct an altitude from A to BC in △ABC and prove two right triangles are congruent.

Hint
HL or SAS after introducing the altitude.
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