Triangles on the Same Base and Between the Same Parallels

Triangles on the same base and between the same parallels are explained here in this article using the theorem with a solved example.

Theorem:

Two triangles on the same base and between the same parallels are equal in area.

Proof

Consider a parallelogram ABCD and AC to be its diagonals, as shown in the figure.

Triangles on the Same Base and Between the Same Parallels - 1

Assume that AN is perpendicular to DC.

We know that the diagonal of a parallelogram divides the parallelogram into two congruent triangles; we can write,

∆ ADC ≅ ∆ CBA

If the triangles are congruent, they have equal areas.

Thus, the area of triangle ADC is equal to the area of a triangle CBA.

(i.e) ar (ADC) = ar (CBA)

Therefore, we can write

ar (ADC) = ½ ar (ABCD)

ar (ADC) = ½ (DC×AN)

Therefore, the area of the triangle = ½ × Base DC × Corresponding Altitude AN.

From the above formula, we can observe that two triangles with the same base (or equal bases) and equal areas will have equal corresponding altitudes.

In order to have an equal corresponding altitude, the triangles must lie between the same parallels.

Hence, two triangles on the same base and between the same parallels are equal in area.

Converse of this Theorem

Two triangles having the same base (or equal bases) and equal areas lie between the same parallels.

Example

Now, let us understand this theorem with the help of an example.

Question:

Prove that the median of a triangle divides it into two triangles of equal areas.

Solution:

Consider a triangle ABC and AD be the one of its diagonals.

Triangles on the Same Base and Between the Same Parallels - 2

Now, we have to prove that

ar (ABD) = ar (ACD).

Now, draw the line AN perpendicular to BC.

Therefore, the area of triangle ABD becomes,

ar(ABD) = ½ × base × altitude (of ∆ ABD)

In a triangle ABD, the base is BD and the altitude is AN.

Therefore, ar(ABD) = ½ (BD×AN)

As BD = CD, the above-mentioned equation can be written as:

ar(ABD) = ½ (CD×AN)

ar(ABD) = ½ × base × altitude (of ∆ ACD)

ar(ABD) = ar(ACD)

Hence, the median of a triangle divides it into two triangles of equal areas, which is proved.

Practice Problems

Solve the following problems.

  1. Prove that the diagonals of a parallelogram divide the parallelogram into four triangles of equal areas.
  2. E is the midpoint of median AD of a triangle ABC, as shown in the figure. prove that ar (ABE) = ar (ACE).

Triangles on the Same Base and Between the Same Parallels - 3

Video Lesson on Triangles

Stay tuned with BYJU’S – The Learning App and learn all Maths-related concepts easily by exploring more videos.

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