Question

У паралелограм АВСD точка М-середина сторони ВС, а N- середина CD Доведіть, що прямі АМ та AN ділять діагональ ВD на три рiннi части. Допоможіть будь ласка !! Срочно треба

Answer 1

Ответ:

Объяснение:У паралелограм АВСD точка М-середина сторони ВС, а N- середина CD Доведіть, що прямі АМ та AN ділять діагональ ВD на три рiннi части. Допоможіть будь ласка !!

Answer 2

Parallelogram ABCD and Points M and N

To prove that the lines AM and AN divide the diagonal BD into three equal parts, we can use the properties of a parallelogram and the midpoint theorem.

Properties of a Parallelogram

A parallelogram is a quadrilateral with opposite sides that are parallel and equal in length. In parallelogram ABCD, we have the following properties:

1. Opposite sides are parallel: AB || CD and BC || AD. 2. Opposite sides are equal in length: AB = CD and BC = AD. 3. Opposite angles are equal: ∠A = ∠C and ∠B = ∠D. 4. Consecutive angles are supplementary: ∠A + ∠B = 180° and ∠B + ∠C = 180°.

Midpoint Theorem

The midpoint theorem states that if a line segment has a midpoint, then it divides the line segment into two equal parts. In other words, if M is the midpoint of segment BC, then BM = MC.

Proof

Given that M is the midpoint of side BC and N is the midpoint of side CD, we want to prove that lines AM and AN divide diagonal BD into three equal parts.

To prove this, we can use the midpoint theorem and the properties of a parallelogram.

1. Since M is the midpoint of BC, we have BM = MC. 2. Similarly, since N is the midpoint of CD, we have CN = ND.

Now, let's consider the line segment BD. We want to show that it is divided into three equal parts.

3. From property 2 of a parallelogram, we know that AB = CD. Therefore, AD = BC. 4. Using property 4 of a parallelogram, we have ∠A + ∠B = 180°. 5. Since AM is a straight line, we have ∠AMB + ∠BMD = 180°. 6. From property 3 of a parallelogram, we know that ∠A = ∠C. Therefore, ∠AMB = ∠CMD. 7. Combining equations 5 and 6, we have ∠CMD + ∠BMD = 180°. 8. Since ∠CMD and ∠BMD are consecutive angles, they are supplementary. 9. Therefore, ∠CMD = ∠BMD = 90°.

Now, let's consider the line segment BM.

10. From the midpoint theorem, we know that BM = MC. 11. From equation 9, we know that ∠CMD = 90°. 12. Therefore, triangle CMD is a right triangle with CM as the hypotenuse. 13. Since BM = MC and ∠BMD = 90°, triangle BMD is also a right triangle with BM as the hypotenuse. 14. By the Pythagorean theorem, we have BD^2 = BM^2 + MD^2.

Similarly, let's consider the line segment MD.

15. From the midpoint theorem, we know that CN = ND. 16. From equation 9, we know that ∠CMD = 90°. 17. Therefore, triangle CMD is a right triangle with CD as the hypotenuse. 18. Since CN = ND and ∠CMD = 90°, triangle CND is also a right triangle with CN as the hypotenuse. 19. By the Pythagorean theorem, we have CD^2 = CN^2 + ND^2.

From equations 14 and 19, we have:

BD^2 = BM^2 + MD^2 CD^2 = CN^2 + ND^2

20. Since AB = CD (property 2 of a parallelogram), we can substitute CD with AB in equation 19: AB^2 = CN^2 + ND^2.

Now, let's consider the line segment AN.

21. From the midpoint theorem, we know that CN = ND. 22. From equation 20, we have AB^2 = CN^2 + ND^2. 23. Therefore, triangle ANC is a right triangle with AN as the hypotenuse. 24. Since CN = ND and ∠CND = 90°, triangle CND is also a right triangle with CN as the hypotenuse. 25. By the Pythagorean theorem, we have AC^2 = CN^2 + ND^2.

From equations 20 and 25, we have:

AB^2 = CN^2 + ND^2 AC^2 = CN^2 + ND^2

26. Since AB = AC (property 2 of a parallelogram), we can substitute AC with AB in equation 25: AB^2 = CN^2 + ND^2.

From equations 20 and 26, we have:

AB^2 = CN^2 + ND^2 AB^2 = AB^2

27. Therefore, CN^2 + ND^2 = AB^2.

From equation 27, we can conclude that CN^2 + ND^2 = AB^2, which means that lines AM and AN divide diagonal BD into three equal parts.

Note: The proof provided above is a logical deduction based on the properties of a parallelogram and the midpoint theorem. It does not require any external sources as it relies on basic geometric principles.