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Lines and Angles | Deck 2 of 2

Perpendicular Bisector Theorem and Proof Skills

Deck 2 of 2: Triangle Congruence Meets Classical Geometry

In this deck:

  • Prove the Perpendicular Bisector Theorem (both directions)
  • Understand biconditional theorems
  • Connect all four angle and line theorems
Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

What You Will Learn This Deck

By the end of this deck, you should be able to:

  1. Prove the Perpendicular Bisector Theorem — both directions
  2. Construct logical proofs citing definitions, postulates, and theorems

From Deck 1 (as tools): Vertical Angles Theorem, Corresponding Angles Postulate, Alternate Interior Angles Theorem

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Observing Equidistance on a Perpendicular Bisector

Segment AB with perpendicular bisector through midpoint M, point P on bisector with equal distances PA and PB shown

Pick any point P on the bisector. Measure PA and PB. What do you find?

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Biconditional Theorems: Two Proofs Required

The Perpendicular Bisector Theorem is a biconditional:

"A point lies on the perpendicular bisector of if and only if it is equidistant from A and B."

"If and only if" = two separate statements:

  • Forward: If P is on the perpendicular bisector →
  • Converse: If → P is on the perpendicular bisector

Each direction requires its own complete proof.

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Setup: Segment, Midpoint, and Bisector

Definitions established for both proofs:

  • : given segment, endpoints A and B
  • M: midpoint of , so
  • : perpendicular bisector — line through M with
  • P: any point on , so
  • PM: shared side (reflexive property)

Three key facts emerge: , ,

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Forward Proof: On Bisector → Equidistant (SAS)

Triangle PMA and PMB with SAS congruence marks: AM=MB, right angle marks, PM=PM

Statement Reason
Definition of midpoint
Definition of perpendicular
Reflexive Property
SAS (side, angle, side)
CPCTC
Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Check: Identify the Three SAS Congruences

Try this before the next slide.

In the forward proof, SAS requires three congruences. List them:

  1. First side: ____
  2. Included angle: ____
  3. Second side: ____

Name the reason (definition, postulate, or property) for each.

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Answer: The Three Congruences for SAS

  • Side: (midpoint)
  • Angle: (perpendicular)
  • Side: (reflexive)

Together: SAS → .

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Converse Proof: Part 1 — Establish SSS Congruence

Given: ; M is the midpoint of
To Prove: P lies on the perpendicular bisector of

Statement Reason
Given
Definition of midpoint
Reflexive Property
SSS (three pairs of equal sides)
CPCTC
Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Converse Proof: Deriving the Right Angle

From Part 1: , so

Statement Reason
Linear Pair Postulate
Substitution ()
Division Property of Equality
Definition of perpendicular
P lies on the perpendicular bisector Definition of perpendicular bisector
Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Why SSS Here, Not SAS?

The two proofs use different congruence criteria. Here is why:

Forward (→) Converse (←)
Given P on bisector → perpendicular angle known PA = PB → sides known, angle unknown
What we know Two sides + included angle Three sides
Criterion SAS SSS
Goal Derive PA = PB Derive 90° angle at M

The given information determines which congruence criterion to use.

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Check: Why SSS, Not SAS, for the Converse?

Think before reading the answer below.

Question: In the converse proof, why can't we use SAS?

Answer: PA = PB gives sides only — no angle at M. Determining that right angle is the goal; using it as a premise would be circular. SSS avoids needing any angle upfront.

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Four Theorems: How They Connect

Theorem dependency chain: Linear Pair Postulate feeds Vertical Angles Theorem and Linear Pair Supplementary; Corresponding Angles Postulate feeds Alternate Interior Angles Theorem; Triangle Congruence feeds Perpendicular Bisector Theorem

Each theorem becomes a tool for the next.

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Multi-Step Proof: Co-Interior Angles Are Supplementary

Given: , ∠3 and ∠5 are co-interior (same-side interior)
To Prove: ∠3 and ∠5 are supplementary

Statement Reason
Corresponding Angles Postulate ()
Definition of congruent angles
∠5 and ∠7 form a linear pair Definition of linear pair
Linear Pair Postulate
Substitution (step 2 into step 4)
∠3 and ∠5 are supplementary Definition of supplementary angles

This proof uses results from Deck 1 as tools.

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

The Equidistant Far Point: Not Just Nearby

, , on the bisector :

⚠️ The theorem holds for ALL points on the bisector — even very far ones.

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Proof-Writing Strategy: A Five-Step Framework

When approaching any proof:

  1. Diagram and label — draw and name all parts
  2. State given and to prove — before writing anything
  3. Inventory tools — which definitions and theorems apply?
  4. Forward and backward — from given; from conclusion
  5. Connect — find where the chains meet
Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

Key Takeaways: All Four Theorems

Vertical Angles: congruent — linear pairs + algebra
Corresponding Angles: congruent — postulate (parallel required)
Alternate Interior: congruent — corresponding + vertical
Perpendicular Bisector: biconditional — SAS forward, SSS converse

⚠️ Two proofs required · No circular reasoning · Parallel lines required

Grade 9 Geometry | HSG.CO.C.9
Lines and Angles | Deck 2 of 2

What Comes Next: Triangle Theorems

In HSG.CO.C.10 you will:

  • Prove the Triangle Angle Sum Theorem using alternate interior angles
  • Prove the Isosceles Triangle Theorem using the Perpendicular Bisector Theorem
  • Extend to parallelogram properties in HSG.CO.C.11
Grade 9 Geometry | HSG.CO.C.9