If $I(1,0)$ is the center of incircle of triangle ABC,the equation of BI is $x-1=0$ and the equation of CI is $x-y-1=0$,then angle BAC is
(A)$\frac{\pi}{4}$
(B)$\frac{\pi}{3}$
(C)$\frac{\pi}{2}$
(D)$\frac{2\pi}{3}$
I found angle $BIC=45^\circ$,but could not find angle BAC because in case of circumcircle,the condition of angle in arc is half of angle at circumcenter is true but not in case of incircle.How can i find this angle,can some guide me?
BI and CI are bisectors of B and C . Look at the triangle BIC : B/2 + BIC + C/2 =
180
The angle between BI and CI (BIC) = 135
B/2 + 135 + C/2 = 180
B + C = 90
In the triangle ABC : A + B + C = 180
A = 90