Show that $A_{1},B_{1},C_{1}$ are on a straight line which is perpendicular to $OI$ Triangle $ABC$ has a circumcircle $(O)$ and a incircle $(I)$. The external bisectors of $\angle A, \angle B,\angle C$ cut $BC,CA,AB$ at $A_{1},B_{1},C_{1}$. Show that $A_{1},B_{1},C_{1}$ are on a straight line which is perpendicular to $OI$.
I think that, first we prove $A_{1},B_{1},C_{1}$ are on a straight line by Menelaus theorem. But how to prove this line is perpendicular to $OI$????
Can anyone help me please? Thank all of you.


PART ONE: Let us first prove that points $A_1$, $B_1$ and $C_1$ are collinear. 
By applying law of sines to triangle $\triangle ACC_1$:
$${AC_1 \over AC}={\sin\angle ACC_1 \over \sin\angle AC_1C}={\sin(90^\circ-\frac\gamma2) \over \sin\angle AC_1C}\tag{1}$$
By applying law of sines to triangle $\triangle BCC_1$:
$${BC_1 \over BC}={\sin\angle BCC_1 \over \sin\angle AC_1C}={\sin(90^\circ+\frac\gamma2) \over \sin\angle AC_1C}\tag{2}$$
Notice that $\sin(90^\circ-\frac\gamma2)=\sin(90^\circ+\frac\gamma2)$. From (1) and (2) it is obvious that: 
$${AC_1 \over AC}={BC_1 \over BC}$$
$${AC_1 \over BC_1}={b \over a}\tag{3}$$
BTW, this simple relation can be obtianed in a dozen of different ways, I just quoted the first that came to my mind. 
In exactly the same way you can show that:
$${BA_1 \over CA_1}={c \over b},\quad {CB_1 \over AB_1}={a \over c}\tag{4}$$
From (3) and (4):
$${AC_1 \over BC_1}\times{BA_1 \over CA_1}\times{CB_1 \over AB_1}=1$$
...so by Menelaus's theorem, points $A_1$, $B_1$ and $C_1$ are collinear.
PART TWO: Let us now prove that $OI\bot A_1B_1C_1$ (a pretty amazing property, at least to me :)
Notice the shortest side of triangle $ABC$. In our case that is, for example, $AC$. Pick points $C'\in BC$ and $A'\in AB$ such that $AC=CC'=AA'=b$. 

LEMMA: Lines $OI$ and $A'C'$ are perpendicular!   
The fact that $OI\bot A'C'$ is actually well know and you can find several different proofs here. Ignore the first post in the the thread because it is tied to a particular value of angle $\angle B$. Just skip it and focus on a general statement of Darij Grinberg (third post in the thread). His proof is not the simplest one and you should scroll down a little bit and check Yptsoi's short and very ellegant answer. The last proof in the same thread is also very interesting. 
The same problem is discussed in several other places on the web  [1][2].

Now, let us prove that triangles $\triangle A_1BC_1$ and $\triangle C'BA'$ are similar. Let us start from (3):
$${AC_1 \over BC_1}={b \over a}$$
$${BC_1 - AB\over BC_1}={b \over a}$$
$$1-{c \over BC_1}={b \over a}$$
$$BC_1={ac \over a-b}\tag{5}$$
Using the same approach:
$$BA_1={ac \over c-b}\tag{6}$$
It is also obvious that:
$$BA'=c-b\tag{7}$$
$$BC'=a-b\tag{8}$$
From (5), (6), (7) and (8):
$$\frac{BC_1}{BA'}=\frac{ac}{(a-b)(c-b)}=\frac{BA_1}{BC}\tag{9}$$
Triangles $\triangle A_1BC_1$ and $\triangle C'BA'$ also share the same angle $B$ so by (9) they are proved to be similar. 
This simply means that $A'C' \parallel A_1C_1$ (red lines in the picture). Our LEMMA states that $OI\bot A'C'$ and therefore $OI\bot A_1B_1C_1$.
