In this blog, we will discuss triangles on the same base and between the same parallels. We saw that the parallelograms on the same base and between the same parallel lines are equal in area.

Now let us see this relation in triangles also which are on the same base and between the same parallels.

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

Now for this theorem, it is given that, ABCD is a parallelogram whose one of the diagonals is AC. Let us perform a construction.

Let AN?DC.

We have to prove that area of triangle ADC = area of triangle CBA.

Let us start the proof

In ?ADC and ?CBA

AD = CB ( as opposite sides of parallelogram)

AC = AC (Common side)

DC = BA (as opposite sides of parallelogram)

?ADC is congruent to ?CBA by side side side. And we know that areas of congruent triangles are equal. So  area (ADC) = area (CBA).

What is the area of ?ADC?

The area of DADC = � base x height.

i.e.                          = DC x AL.

And the area of parallelogram ABCD is = base x altitude

                                                               = DC x AL

Therefore, area (ADC) =1/2 area (ABCD)

In other words, the area of a triangle is half the product of its base and the corresponding altitude. So we can see that two triangles with the same base and equal areas will have equal corresponding altitudes.

Theorem: Two triangles having the same base and equal areas lie between the same parallels.

Here we are given Triangles ABC and DCB have the same base CB. And the area of triangle ABC is equal to the area of triangle DCB

And we have to prove that CB is parallel to AD. As construction Let us draw altitude DL and AP on CB.

Let us start the proof: In ?ABC and ?DCB

            ar (ABC) = ar (DCB)                                 (Given)

     i.e. 1/2CB x DL = � CB x AP

                         DL = AP

This implies the distance between lines AD and CB is equal which shows that lines AD and CB are ll.

Let us take one more example.

Example: Show that a median of a triangle divides it into two triangles of equal areas.

We are given here that ABC is a triangle and let AD be one of its medians

We have to prove that: area (ABD) = area (ACD). Since the formula for area involves altitude, let us draw  AN perpendicular to BC.

Let us start the proof

In ?ABD

Area (ABD) = half  base � altitude

Which is equal to half  BD x AN

This is equal to half  CD x AN (As BD = CD)

But CD is the base of the triangle ACD and AN is the height.

So above expression can be written as

1/2 � base � altitude for ?ACD

Which is the same as area (ACD)

Therefore we can say that area of triangle ABD is equal to the area ACD

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