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Geometry Theorems SAA Axiom Proof
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Theorem (SAA Axiom)
Parallelograms Standing on the Same Base
Statement / Question: Prove that parallelograms standing on the same base and between the same parallel lines are equal in area using the Side-Angle-Angle (SAA) congruence axiom.
Given
▱ABCD & ▱ABEF stand on base AB between FC ∥ AB.
To Prove
Area of ▱ABCD = Area of ▱ABEF
Theoretical Proof (Area Subtraction Method)
| Statements | Reasons |
|---|---|
|
1. In △ADF and △BCE: (i) AF = BE (S) (ii) ∠AFD = ∠BEC (A) (iii) ∠ADF = ∠BCE (A) |
1. (i) Opp. sides of ▱ABEF (ii) Corr. angles (AF ∥ BE) (iii) Corr. angles (AD ∥ BC) |
| 2. △ADF ≅ △BCE | 2. By SAA congruence axiom |
| 3. Area(△ADF) = Area(△BCE) | 3. Congruent triangles are equal in area |
| 4. Quad ABCF – Area(△BCE) = Quad ABCF – Area(△ADF) | 4. Subtracting St. (3) from Quad ABCF on both sides |
| 5. Area(▱ABEF) = Area(▱ABCD) | 5. Remaining areas are equal (from St. 4) |
✓ PROVED
Figure: Case 1 (Standard Overlapping)
Base: AB
Case 1: Parallelograms ▱ABEF & ▱ABCD with overlapping area.