Question

 Середины оснований трапеции соединены отрезком.       Докажите, что полученные две трапеции равновелики.

Answer 1

I will answer in detail. The midpoints of the bases of a trapezoid are connected by a segment. Prove that the resulting two trapezoids are equal in area.

To prove this, we can use the following steps:

1. Let ABCD be a trapezoid with bases BC and AD, and let M and N be the midpoints of BC and AD respectively. Draw the segment MN, which divides the trapezoid into two smaller trapezoids ABMN and NMCD. 2. We need to show that the areas of ABMN and NMCD are equal, that is, S(ABMN) = S(NMCD). 3. To do this, we can use the formula for the area of a trapezoid: S = (a + b)h / 2, where a and b are the lengths of the bases and h is the height. 4. In the trapezoid ABCD, we have a = BC, b = AD, and h = MN. Therefore, S(ABCD) = (BC + AD)MN / 2. 5. In the trapezoid ABMN, we have a = BM, b = AN, and h = MN. Since M and N are the midpoints of BC and AD, we have BM = MC = BC / 2 and AN = ND = AD / 2. Therefore, S(ABMN) = (BM + AN)MN / 2 = (BC / 2 + AD / 2)MN / 2 = (BC + AD)MN / 4. 6. Similarly, in the trapezoid NMCD, we have a = MC, b = ND, and h = MN. Therefore, S(NMCD) = (MC + ND)MN / 2 = (BC / 2 + AD / 2)MN / 2 = (BC + AD)MN / 4. 7. Comparing the expressions for S(ABMN) and S(NMCD), we see that they are equal. Hence, S(ABMN) = S(NMCD), which proves that the two trapezoids are equal in area.

Here is a diagram of the trapezoid and the segment MN:

![Trapezoid](https://i.imgur.com/0jZ2L8w.png