<s>[INST] How find the $\angle B$ 
In $\Delta ABC$ such $I$ is incentre,and $$\angle A=80^{0},AI+IB=BC$$,
find the $\angle B$
my idea:let $AB=c,AC=b,BC=a$ then we have
$$\dfrac{AI}{ID}=\dfrac{AB}{AD}=\dfrac{AC}{DC}=\dfrac{AB+AC}{AD+DC}=\dfrac{b+c}{a}$$
and
$$AD^2=AB\cdot AC-BD\cdot DC=bc-BD\cdot DC$$
$$BD=\dfrac{ac}{b+c},DC=\dfrac{ab}{b+c}$$
$$\Longrightarrow AD=\sqrt{bc-\dfrac{a^2bc}{(b+c)^2}}$$
so
$$AI=\dfrac{b+c}{a+b+c}AD$$
and 
$$BI=\dfrac{a+c}{a+b+c}BE$$
so
$$AI+BI=BC\Longleftrightarrow \dfrac{b+c}{a+b+c}AD+\dfrac{a+c}{a+b+c}BE=a$$
 then I have ugly,and can't work,Thank you someone can have other methods.
 [/INST] Hint: Apply sin rule to $BIC$, get $BC$ in terms of $BI$.
Hint: Apply sin rule to $BIA$, get $AI$ in terms of $BI$.
Now substitute these into $AI + BI = BC$, you will get a trigonometric equation in terms of $\angle IBC$. Solve it to determined that $\angle IBC = 20^\circ$.
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