IFE,F,G,H ARE RESPECTIVELY THE MID POINTS OF THE SIDE OF A PARALLELOGRAM ABCD SHOW THAT AREA OF EFGH IS EQUAL TO HALF AREA OF ABCD.

To prove that the area of EFGH is equal to half the area of ABCD, we need to use the concept of the triangle area.

Step 1: Since I, F, and E are the midpoints of the sides of parallelogram ABCD, we can see that triangles AEF, BED, CFI, and DGH are all congruent. This is because the midpoints divide the sides into two equal segments, so these triangles share the same base and height.

Step 2: Let's take triangle AEF as an example. It has the same base as triangle ABC, which is EF, and the same height as triangle ABC, which is the perpendicular distance between EF and AB. Therefore, the area of triangle AEF is half the area of triangle ABC.

Step 3: By applying the same reasoning to triangles BED, CFI, and DGH, we can conclude that each of these triangles has an area that is half the area of ABC.

Step 4: Since triangles AEF, BED, CFI, and DGH cover the entire area within EFGH, the area of each of these triangles combined is equal to the area of EFGH.

Step 5: Therefore, the area of EFGH is equal to the sum of the areas of triangles AEF, BED, CFI, and DGH, which is equal to the sum of half the area of ABC.

Step 6: Mathematically, we can represent this as follows:
Area of EFGH = (Area of AEF + Area of BED + Area of CFI + Area of DGH)
= (1/2 * Area of ABC + 1/2 * Area of ABC + 1/2 * Area of ABC + 1/2 * Area of ABC)
= 2 * (1/2 * Area of ABC)
= 1/2 * Area of ABC.

Step 7: Therefore, we have proven that the area of EFGH is equal to half the area of ABCD.

To prove that the area of the quadrilateral EFGH is equal to half the area of the parallelogram ABCD, we can use the fact that the midpoints of the sides of a parallelogram divide it into four congruent triangles.

Here is one way to prove it:

1. Draw the parallelogram ABCD with the midpoints I, F, G, and H labeled as stated.

2. Connect the midpoints I and G to form the diagonal IG. Similarly, connect the midpoints F and H to form the diagonal FH.

3. Now, we have divided the parallelogram ABCD into two congruent triangles, AIF and CID, and two congruent triangles GBH and CFD.

4. Since the diagonals of a parallelogram bisect each other, we can say that IG is the midline of the triangle ACD and FH is the midline of the triangle BCD.

5. The midline of a triangle divides it into two triangles with equal areas. Therefore, the triangles AIF and CID have equal areas, and the triangles GBH and CFD have equal areas.

6. The quadrilateral EFGH is formed by the combination of these four triangles. Since the triangles AIF and CID have equal areas, and the triangles GBH and CFD have equal areas, the quadrilateral EFGH is formed by two pairs of congruent triangles.

7. Thus, we can conclude that the area of the quadrilateral EFGH is equal to half the area of the parallelogram ABCD.

Overall, by dividing the parallelogram ABCD into congruent triangles using the midpoints of its sides, we can see that the area of the quadrilateral EFGH is equal to half the area of the parallelogram.