How find the $\angle B$

Question

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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.

Well you have a incentre where all three collinear lines go through. Then it follows that the line segments are angle bisectors. — user60887, Sep 20 '13 at 04:10
Well angles CAD and DAB are 40 degrees a piece. Do you know what an incentre is? — user60887, Sep 20 '13 at 04:18
Of course I know what is the Angle bisector, know the two Angle of 40 degrees have what effect — , Sep 20 '13 at 04:29
I used the Angle bisector theorem, what do you think I know what is inside!! — , Sep 20 '13 at 04:31
$\angle B=40°$, found through manual gradient descent with Cinderella. — MvG, Sep 20 '13 at 13:21
There is a typo in your first equation. When you say $AD$, you mean $BD$. — Calvin Lin, Sep 20 '13 at 15:43

score 2 · Accepted Answer · answered Sep 20 '13 at 15:51

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 $\angle IBC = 20^\circ$.

How find the $\angle B$

1 Answers1