1. ## parallelogram

Prove that parallelograms on the same base and betweem the same parallels are equal in area .

2. Hello thereddevils
Originally Posted by thereddevils
Prove that parallelograms on the same base and betweem the same parallels are equal in area .
Divide the parallelogram into two equal triangles, using a diagonal. Then the area of the parallelogram = twice the area of one of the triangles. Area of triangle = half base times height. So?

3. Originally Posted by thereddevils
Prove that parallelograms on the same base and betweem the same parallels are equal in area .
Hi thereddevils,

Draw parallelogram ABCD with AB being the top side. Extend AB to E such that BE = DC.

Connect D to B and C to E.

You have just constructed another parallelogram BECD with same base as ABCD and between the same parallels, as required.

In parallelogram ABCD with diagonal BD, we can prove that the two triangles ABC and CDB are congruent.

We also have sufficient evidence to prove that triangles CDB and BEC are congruent.

Now all three triangles are congruent to each other. They have equal areas.

4. oh thansk a lot , Grandad and masters . i was just a little confused with the meaning of between the parallels .

5. Hello everyone
Originally Posted by masters
Hi thereddevils,

Draw parallelogram ABCD with AB being the top side. Extend AB to E such that BE = DC.

Connect D to B and C to E.

You have just constructed another parallelogram BECD with same base as ABCD and between the same parallels, as required.

In parallelogram ABCD with diagonal BD, we can prove that the two triangles ABC and CDB are congruent.

We also have sufficient evidence to prove that triangles CDB and BEC are congruent.

Now all three triangles are congruent to each other. They have equal areas.
While this is true, it's a very special case. You should really consider two general parallelograms on the same base and between the same parallels, as I have indicated in the attached drawing. I outlined the proof in my first posting. Here are a few more details.

In the diagram, $ABCD$ and $CDE F$ are any two parallelograms on the base CD, with the points $A,B, E,F$ in a straight line.

Construct the diagonals $BD$ and $CE$. Triangles $ABD$ and $CDB$ can be easily proven congruent; as can triangles $DEC$ and $FCE$.

Drop perpendiculars from $B, E$ to meet $CD$ at points $X, Y$. $BEYX$ is then a rectangle. Therefore $BX = EY$.

Area $ABCD = 2\times$ area $\triangle CDB = CD \times BX = CD \times EY =$ area $CDE F$.