Triangles on the Same Base and Between the Same Parallels
Statement / Question: Prove that triangles standing on the same base and lying between the same parallel lines are equal in area.
Given
△ABC & △DBC on base BC between EF ∥ BC.
To Prove
Area(△ABC) = Area(△DBC)
Construction
Draw perpendicular MN ⊥ BC from line EF to base BC.
Theoretical Proof
| S.N. | Statements | Reasons |
|---|---|---|
| 1 | Area(△ABC) = ½ × BC × MN | Area of triangle = ½ × base × height |
| 2 | Area(△DBC) = ½ × BC × MN | Area of triangle = ½ × base × height |
| 3 | Area(△ABC) = Area(△DBC) | From statements 1 and 2 (Equating both areas). |
Triangles Standing on Equal Bases Between the Same Parallels
Statement / Question: Prove that triangles standing on equal bases and lying between the same parallel lines are equal in area.
Given
△ABC & △DEF on equal bases BC = EF between AD ∥ BF.
To Prove
Area(△ABC) = Area(△DEF)
Construction
Draw perpendiculars MN ⊥ BC & PQ ⊥ EF.
Theoretical Proof
| S.N. | Statements | Reasons |
|---|---|---|
| 1 | MN = PQ = h | Perpendicular distance between parallel lines (AD ∥ BF) is equal. |
| 2 | Area(△ABC) = ½ × BC × h | Area of triangle = ½ × base × height |
| 3 | Area(△DEF) = ½ × EF × h | Area of triangle = ½ × base × height |
| 4 | Area(△ABC) = Area(△DEF) | From st. 2 & 3, since BC = EF & heights are equal. |
3. Dynamic Experimental Verification Sandbox
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