Question

Доказать, что если в треугольнике (a-b)/a = 1-2cosC, то треугольник равнобедренный.

Answer 1

Представим данное равенство в виде $\dfrac{a}{a}- \dfrac{b}{a} =1-2\cos C$ или $1-\dfrac{b}{a} =1-2\cos C$, откуда $2\cos C=\dfrac{b}{a}$ .
Согласно теореме косинусов имеем $c^2=a^2+b^2-2ab\cos C$ .
Подставим значение $2\cos C=\dfrac{b}{a}$, получим $c^2=a^2+b^2-b^2$ или $c^2=a^2$, откуда, так как c>0 и a>0, найдем $c=a$

Таким образом, если в треугольнике $\dfrac{a-b}{a} =1-2\cos C$, то треугольник равнобедренный.

Что и требовалось доказать.

Answer 2

Proof that if (a-b)/a = 1-2cosC, then the triangle is isosceles

To prove that if (a-b)/a = 1-2cosC, then the triangle is isosceles, we need to use trigonometric identities and properties of triangles.

Let's start by assuming that we have a triangle ABC, where side AB has length a, side BC has length b, and angle C is opposite side AB.

We are given the equation (a-b)/a = 1-2cosC. We can simplify this equation as follows:

(a-b)/a = 1-2cosC (a-b) = a(1-2cosC) a - b = a - 2acosC b = 2acosC

Now, let's consider the Law of Cosines, which states that in a triangle with sides a, b, and c, and angle C opposite side c, the following equation holds:

c^2 = a^2 + b^2 - 2abcosC

In our case, we have b = 2acosC. Substituting this into the Law of Cosines equation, we get:

c^2 = a^2 + (2acosC)^2 - 2a(2acosC)cosC c^2 = a^2 + 4a^2cos^2C - 4a^2cos^2C c^2 = a^2

From this equation, we can see that c^2 = a^2, which means that side AC is equal in length to side AB. Therefore, the triangle ABC is an isosceles triangle.

In conclusion, if (a-b)/a = 1-2cosC, then the triangle ABC is isosceles.