Question

Отрезки AB и CD пересикаются в т M которая является серединой этих отрезков докажите что AC=CD. спасибо за помощь

Answer 1

Т.к. М-середина обоих отрезков, то АМ=МВ=СМ=МД. Т.е. в треугольнике АМС и треугольнике ДМВ равны попарно стороны АМ=ДМ и СМ=МВ. А угол АМС= углу ДМС как вертикальные углы. => треугольник АСМ = треугольнику ДМВ по 1 признаку равенства треуглльников. А раз треугольники равны, то и все стороны в них равны => АС=ДВ.

Answer 2

Proof that AC = CD

To prove that AC = CD, we can use the fact that point M is the midpoint of both segments AB and CD. Let's break down the proof step by step:

1. Given: Segments AB and CD intersect at point M, which is the midpoint of both segments.

2. We know that M is the midpoint of segment AB, so we can write: AM = MB. 3. Similarly, M is the midpoint of segment CD, so we can write: CM = MD. 4. Now, let's consider the triangle AMC. Since M is the midpoint of AB, we have: AM = MB. 5. By the transitive property of equality, we can say that AM = MB = AC.

6. Similarly, let's consider the triangle CMD. Since M is the midpoint of CD, we have: CM = MD. 7. By the transitive property of equality, we can say that CM = MD = CD.

8. Combining the results from steps 5 and 7, we have: AC = AM + CM = MB + MD = CD.

Therefore, we have proved that AC = CD.

I hope this explanation helps! Let me know if you have any further questions.