Question

в треугольнике KMN угол K= углу N. на высоте М Н отмечена точка D.докажите что треугольник KDN равнобедренный....пожалуйста решите очень надо

Answer 1

в треугольнике KDN DH-является высотой и медианой, а => по признаку р/б треугольников треуг.KDN-р/б

Answer 2

Given Information:

We are given a triangle KMN, where angle K is equal to angle N. Point D is marked on the altitude MN.

To Prove:

We need to prove that triangle KDN is isosceles (or, in other words, that triangle KDN is an isosceles triangle).

Proof:

To prove that triangle KDN is isosceles, we need to show that KD = DN.

Since angle K is equal to angle N, we can conclude that triangle KMN is an isosceles triangle, with KM = MN.

Now, let's consider triangle KMD. Since KM = MD (as triangle KMN is isosceles), we have two sides of triangle KMD equal.

Additionally, we know that angle K = angle N. This means that angle KMD = angle NMD, as they are corresponding angles.

By the Side-Angle-Side (SAS) congruence criterion, we can conclude that triangle KMD is congruent to triangle NMD.

Therefore, KD = DN, and triangle KDN is an isosceles triangle.

Conclusion:

We have proven that triangle KDN is an isosceles triangle, with KD = DN.