In an isosceles triangle $△ABC$ with $AB=AC$, $I$ is the incentre such that $BC=AC+AI$. Find $∠A$.
My Try:
We have $BC=a$, $AB=c$, $AC=b$, and $AI=AM-IM$, where $AM$ is the altitude to base of the triangle. Now
$$AI=\frac{a}{2}\tan B-\frac{a}{2} \tan\left(\frac{B}{2}\right),$$
hence using $BC=AC+AI$ we get
$$a-b=\frac{a}{2}\tan B-\frac{a}{2} \tan\left(\frac{B}{2}\right).$$
How can we proceed further?
Because $AB = AC$, then $$\sin A = \sin(π - 2B) = \sin 2B.$$ Note that $0 < B < \dfrac{π}{2}$, then$$ \tan \frac{B}{2} = \frac{\sin B}{1 + \cos B}. $$ Thus\begin{align*} &\mathrel{\phantom{\Longleftrightarrow}} a - b = \frac{a}{2} \tan B - \frac{a}{2} \tan \frac{B}{2}\\ &\Longleftrightarrow \sin 2B - \sin B = \frac{1}{2} \sin 2B \tan B - \frac{1}{2} \sin 2B \tan \frac{B}{2}\\ &\Longleftrightarrow \sin B (2\cos B - 1) = \sin B \left(\sin B - \cos B \tan \frac{B}{2} \right)\\ &\Longleftrightarrow 2\cos B - 1 = \sin B - \cos B \tan \frac{B}{2}\\ &\Longleftrightarrow (2\cos B - 1)(1 + \cos B) = \sin B (1 + \cos B) - \cos B \sin B\\ &\Longleftrightarrow 2\cos^2 B + \cos B - 1 = \sin B\\ &\Longrightarrow (2\cos^2 B + \cos B - 1)^2 = 1 - \cos^2 B\\ &\Longleftrightarrow (1 + \cos B)^2 (2\cos B - 1)^2 = (1 + \cos B)(1 - \cos B)\\ &\Longleftrightarrow (1 + \cos B)(2\cos B - 1)^2 = 1 - \cos B \quad (\because 0 < B < \frac{π}{2})\\ &\Longleftrightarrow 4\cos^3 B - 2\cos B = 0\\ &\Longleftrightarrow \cos B = \frac{1}{\sqrt{2}}. \quad (\because 0 < B < \frac{π}{2}) \end{align*} Then$$ B = \frac{π}{4} \Longrightarrow A = \frac{π}{2}. $$
A geometric proof:
Suppose circle $AIC$ intersect with $BC$ at point $D \neq C$. Because $$ ∠BAD = ∠BAI + ∠DAI = ∠BAI + ∠DCI = ∠BAI + ∠ABI\\ = π - ∠AIB = π - ∠AIC = π - ∠ADC = ∠BDA, $$ then $BA = BD$. Thus,$$ AI = BC - AB = BC - BD = CD, $$ then$$ ∠ACI = ∠CID\\ \Longrightarrow \frac{1}{2} ∠BAC = ∠CAI = ∠CAD + ∠DAI = ∠ACI + ∠DCI = ∠ACB. $$ Now it is easy to get $∠BAC = \dfrac{π}{4}$.