In the triangle $\triangle ABC$, let $BB_1$ and $CC_1$ be the bisectors of angles $B$ and $C$, $B_1$ lies on $AC$, and $C_1$ lies on $AB$. Let $B_2$ and $C_2$ be the midpoints of ${BB_1}$ and ${CC_1}$ respectively. Show that the lines $C_1B_2$, $B_1C_2$, and $BC$ are concurrent if and only if $\angle A = 60^\circ$.
I have tried writing the equation of ${B_2C_1}$ and ${B_1C_2}$ and ${CB}$ in the complex plane and then let P be the intersection of all 3 but the calculations seem too complicated knowing that the equation of a angle bisector has a real parameter.
Solving this question using vector method. Let A be the Origin $\vec AB= \vec b , \vec AC = \vec c, |\vec BC|=a$; using angle bisector theorem, $$\vec B_1= \frac{b}{a+b} \vec c \qquad \vec C_1= \frac{c}{a+c} \vec b$$ $$\vec B_2= {1\over2}\left(\vec b + \frac{b}{a+b} \vec c \right) \qquad \vec C_2= {1\over2}\left(\vec c + \frac{c}{a+c} \vec b \right) $$ lines $C_1B_2, C_2B_1$ meet at some point $Q$ then it is possible to find real parameters $\lambda$ and $\mu$ such that, $$\vec Q= \vec B_1 + \lambda\left( \vec C_2-\vec B_1\right)=\vec C_1+\mu \left(\vec B_2-\vec C_1\right)$$ putting values $\vec B_1, \vec C_1, \vec C_2, \vec B_2$ in terms of $\vec b$ and $\vec c$ (that we have written earlier) and comparing co-eff. of $\vec b, \vec c$ (according to concept of independency of two non colinear vectors) you will get $$(a-b)\lambda-b\mu+2b=0$$ $$(a-c)\mu-c\lambda+2c=0$$ solving these two. $$\lambda= {2b \over b+c-a}$$
Now, $$\vec Q= \vec B_1 + \lambda\left( \vec C_2-\vec B_1\right)$$ $$= \frac{b}{a+b} \vec c + \lambda \left[{1\over2}\left(\vec c + \frac{c}{a+c} \vec b \right)-\frac{b}{a+b} \vec c\right]$$ as Q is one the line BC, so sum of co-eff. of $\vec c$ and $\vec b$ $= 1$ from this you will get another value of $\lambda$ in terms of $b,c$
$$\lambda= {2(a+c) \over a-b+2c}$$
equating these two values of $\lambda$ you will finally reach here, $$b^2 + c^2 - a^2 = bc$$