Question

Show that diagonals of a parallelogram divides it into 4 triangles of equal area.

Plz give clear explanation for each step & attach the diagram too...

Answer 1

Okay so, first of all let's sort out what is given..

You're given a parallelogram ABCD, Where AC and BD are diagonals which intersect at O. Now you see, you'll be given 4 triangles that are OAB, OBC, OCD and ODA.

Now, let's further see what do you have to prove..

To prove- Area of OAB= Area of  OBC= Area of OCD= Area of ODA.

Constructions- You'll require a corresponding altitude to prove it. So, let's draw BE perpendicular to AC.

Proof- Since, ABCD is a parallelogram, OA=OB=OC=OD. (diagonals of parallelogram bisect each other)

Now, Area of OAB= (Base*Corresponding altitude)/2= OA*BE/2
And, Area of OBC= (Base*Corresponding altitude)/2= OC*BE/2

But as mentioned above, OA=OC
Therefore, Area of OAB=Area of OBC
Similarly,
                 Area of OBC=Area of OCD
And, 
                  Area of OCD=Area ofODA

From all these equations, we prove that all the 4 triangles have equal area. :)