Ratio of two line segments in a special right triangle

Question

So I had this question in my mind for a while.

At first, I tried using a bunch of theorems such as the Law of Sines, the Law of Cosines, and similarity. I got the answer $\frac{a}{a-1}$, but then I realized it was wrong because of a factoring issue.

Answer 1

The triangle $\,ABC\,$ is similar to $\,EDC.\,$ Use similarity of sides to get $$ \frac{r_1}{a+r_1} = \frac{r_2}a = \sin(\alpha). \tag{1} $$ This implies $$ 1-\sin(\alpha) = 1 - \frac{r_2}a = \frac{a-r_2}a. \tag{2} $$ Equality of ratios in equation $(1)$ gives us $$ r_1\,a= (a+r_1)\,r_2 = a\,r_2 + r_1r_2. \tag{3} $$ Rearrange terms and and add $\,a^2\,$ to get $$ a^2 = a^2 + r_1\,a - a\,r_2 - r_1r_2 = (a+r_1)(a-r_2). \tag{4} $$ This implies the equality of ratios $$ \frac{a+r_1}a = \frac{a}{a-r_2}. \tag{5} $$ Combine equality of ratios in equation $(1)$ with equations $(2)$ and $(5)$ to get $$ \frac{r_1}{r_2} = \frac{a+r_1}a= \frac{a}{a-r_2} = \frac1{1-\sin(\alpha)}. \tag{6} $$