Triangles on the same base and between the same parallels

Triangles on the same base and between the same parallels have equal areas. This concept uses the principle of base and height to compare triangles easily. In this guide, we will learn about the theorem, its proof, and related examples for easy understanding of the concept.

Table of Contents

• Theorem on Triangles on the Same Base and Between Same Parallels
• Solved Examples on Triangles on the Same Base and Between the Same Parallels

Theorem on Triangles on the Same Base and Between the Same Parallels

Theorem: Triangles on the same base and between the same parallels have the same area. 

Here, the areas of ∆ABC and ∆ABD are equal. Both triangles have the same base AB and are under the same parallel lines.

Proof:

Let ABC and ABD be two triangles with same base and between same parallel lines.

Draw BF ∥ AD and AE ∥ BC

In ABFD,

AB ∥ DF (given as AB ∥ EF), and BF ∥ AD (by construction)

Hence, ABFD is a parallelogram.

Similarly, ABCE is a parallelogram.

ABFD and ABCE are parallelograms on the same base and between the same parallels.

∴ ar(ABFD) = ar(ABCE) (∵ area of parallelograms on the same base and between the same parallels are equal). --------- (1)

In parallelogram ABFD, since the diagonal divides the parallelogram into two congruent triangles

 ∆ABD ≅  ∆FBD

∴ ar (ABD) = ar (FBD) (∵ area of congruent triangles is equal)

∴ ar (ABFD) = 2 ar (ABD)    ---------- (2)

Similarly, in parallelogram ABCE, ∆ABC ≅ ∆AEC.

∴ ar(ABC) = ar(AEC)

∴ ar (ABCE) = 2 ar (ABC)  --------- (3)

From (1), ar(ABFD) = ar(ABCE)

From (2) and (3), 2 ar(ABD) = 2 ar(ABC).     

⇒ ar(ABD) = ar(ABC)

Hence proved.

NOTE: The converse of the theorem holds, i.e., ΔABC and ΔABD are on the same base BC such that ar(ΔABC) = ar(ΔABD). Then AB ∥ EF

Solved Examples on  Triangles on the Same Base and Between the Same Parallels

Example 1: ΔABC and ΔDBC are on the same base BC. A and D are on the same side of BC such that AD ∥ BC. If BC = 16 cm and the distance between BC and AD is 3 cm, find the area of each triangle.
Solution: Given that ΔABC and ΔDBC are on the same base, BC, and between the same parallel lines.
∴ ar(ABC) = ar(DBC)
ar(ABC) = (½)bh = (½) × 16 × 3 = 24 cm²
∴ The area of each triangle is 24 cm².

Example 2: ΔPQR has base QR = 12 cm and height from P = 8 cm. ΔSQR has the same base QR, and ar(ΔSQR) = ar(ΔPQR). Find the height from S to QR.
Solution: ΔPQR has base QR = 12 cm and height from P = 8 cm.
ΔSQR has the same base QR, and ar(ΔSQR) = ar(ΔPQR).
∴ Both triangles are between the same parallel lines, and PS ∥ QR.
ar (PQR ) = (½) × 12 × 8 = 48 cm²
ar (SQR) =  (½) × 12 × h = 48 cm²
∴ h = 8 cm.
Hence, the height from S to QR is 8 cm.