Question

Дан параллелограмм ABCD. O- точка пересечения диагоналей. Докажите что треугольник AOB=треугольнику COD

Answer 1

Противоположные стороны параллелограмма параллельны и равны:
АВ=СД и АД=ВС
Диагонали параллелограмма точкой пересечения делятся пополам:
АО=СО и ВО=ОД
Следовательно ΔАОВ=ΔСОД по 3 сторонам, ч.т.д.

Answer 2

Proof that triangle AOB is congruent to triangle COD

To prove that triangle AOB is congruent to triangle COD, we can use the fact that they share a common base, AB, and have equal heights. Here's the step-by-step proof:

1. Given: Parallelogram ABCD with O as the point of intersection of the diagonals.

2. We know that the diagonals of a parallelogram bisect each other. Therefore, AO is equal to OC and BO is equal to OD.

3. We also know that the opposite sides of a parallelogram are equal in length. Hence, AB is equal to CD.

4. Triangle AOB and triangle COD share a common base, AB.

5. To prove that triangle AOB is congruent to triangle COD, we need to show that they have equal heights.

6. Let's consider the height of triangle AOB. The height of a triangle is the perpendicular distance between the base and the opposite vertex.

7. Since AB is parallel to CD, the height of triangle AOB is equal to the perpendicular distance between AB and CD.

8. Similarly, the height of triangle COD is equal to the perpendicular distance between AB and CD.

9. Since AB is equal to CD, the perpendicular distances between AB and CD are also equal.

10. Therefore, the heights of triangle AOB and triangle COD are equal.

11. As triangle AOB and triangle COD have a common base and equal heights, they are congruent by the Side-Angle-Side (SAS) congruence criterion.

12. Hence, triangle AOB is congruent to triangle COD.

In conclusion, we have proved that triangle AOB is congruent to triangle COD based on the given parallelogram ABCD with O as the point of intersection of the diagonals.

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