Question

На боковых сторонах равнобедренного треугольника АВС отложены равные отрезки ВМ и ВN, ВD-медиана треугольника. Докажите, что MD=ND

Answer 1

Рассмотрим треугольники and и dmc
они равны по двум сторонам и углу между ними (na=mc, ad=dc, угол a = углу с)
Из равенства треугольников получаем равенство соответствующих элементов, след-но nd=dm

Answer 2

Чертеж и решение во вложении

Answer 3

Given Information

The problem states that on the sides of an isosceles triangle ABC, equal segments VM and VN are marked, and BD is the median of the triangle.

Proof of MD = ND

To prove that MD = ND, we can use the properties of medians in a triangle.

In an isosceles triangle, the median from the vertex angle is also an altitude and a perpendicular bisector. This means that the median BD is also an altitude and a perpendicular bisector, which divides the base AC into two equal parts.

Since BD is a median, it divides the base AC into two equal parts. Therefore, the segments AM and CN are also equal.

Now, let's consider the triangles MBD and NBD. Both triangles share the side BD, and we have established that AM = CN. Since AM = CN, the triangles MBD and NBD share two sides that are equal, and they also share the median BD.

By the Side-Side-Side (SSS) congruence criterion, we can conclude that the triangles MBD and NBD are congruent.

When two triangles are congruent, their corresponding sides are equal. Therefore, MD = ND.

Therefore, MD = ND is proven.

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