prove that the diagonals of any parallelogram divides it into 4 triangles of equal area
Answers
Step-by-step explanation:
Call the parallelogram ABCD and let E be the point of intersection of the diagonals AC and BD.
Use the fact that we know that AC and BD bisect each other, so E is the midpoint of AC and E is the midpoint of BD.
Look at triangles ABE and ADE. Since BE and ED lie in the same line and the triangle share the vertex A, these triangle have the same "height" from A to their base, which is in the line BED. Now area is half times base times height, and here we have heights the same, and also bases are equal (BE = ED since E is the midpoint). So the areas of ABE and ADE are equal.
Likewise, ABE and CBE have the same area since these triangles have the same height from B and their bases are equal (AE = CE).
Likewise, CBE and CDE have the same area since they have the same height from C and BE = DE.
All together, this gives
area(ADE) = area(ABE) = area(CBE) = area(CDE).