Determine the ratio of two angles in a triangle.

[anemone]

Let \(\displaystyle ABC\) be a triangle such that \(\displaystyle \frac{BC}{AB-BC}=\frac{AB+BC}{AC}\).

Determine the ratio \(\displaystyle \frac{\angle A}{\angle C}\).

 

[Opalg]

Let \(\displaystyle ABC\) be a triangle such that \(\displaystyle \frac{BC}{AB-BC}=\frac{AB+BC}{AC}\).

Determine the ratio \(\displaystyle \frac{\angle A}{\angle C}\).

Write $a,\ b,\ c$ for the sides opposite the angles $A,\ B,\ C$ respectively. You are told that $\dfrac a{c-a} = \dfrac{c+a}b$, so that $ab = c^2-a^2$. The cosine rule tells you that $c^2 = a^2+b^2-2ab\cos C$. Combining those equations, you get $ab = b^2 -2ab\cos C$, so that $a = b-2a\cos C$. Now draw a picture.

The base of the isosceles triangle $BCD$ is twice the projection of $BC$ onto the side $CA$, namely $2a\cos C$. Therefore $DA = b-2a\cos C = a$. Thus $BDA$ is also isosceles, and it follows that $\angle C = \angle BDC = 2\angle A$.

 

[anemone]

Write $a,\ b,\ c$ for the sides opposite the angles $A,\ B,\ C$ respectively. You are told that $\dfrac a{c-a} = \dfrac{c+a}b$, so that $ab = c^2-a^2$. The cosine rule tells you that $c^2 = a^2+b^2-2ab\cos C$. Combining those equations, you get $ab = b^2 -2ab\cos C$, so that $a = b-2a\cos C$. Now draw a picture.

The base of the isosceles triangle $BCD$ is twice the projection of $BC$ onto the side $CA$, namely $2a\cos C$. Therefore $DA = b-2a\cos C = a$. Thus $BDA$ is also isosceles, and it follows that $\angle C = \angle BDC = 2\angle A$.

Awesome! I used a totally different method and yours is without a doubt, so much better than mine!

Thanks for showing me that we could actually solve a geometric problem using such a method and to be honest with you, I always love to read how you would approach a problem...

Thank you so much, Opalg!