How do I construct this triangle? Point $O$ is the centre of the circle inscribed in $ABC$, on sides $AC$ and $BC$ points $M$ and $K$, respectively, are taken so that $BK \times AB = BO^2$ and $ AM \times AB = AO^2$. Prove that $M, O$ and $K$ are collinear.
I'm working on this problem here an I'm having a hard time here trying to construct the the given triangle so that I can proceed to solve the problem. I do understand that $OB$ is the geometric mean of $A$ and $B$ and that $AO$ is the geometric mean of $M$ and $B$ but what I don't understand is how to go about constructing these points in the given triangle(because I know how to construct geometric mean).
 A: My religion forbids me from calling the incenter as $O$, so I will use the letter $I$.

Let $\rho_B$ a clockwise rotation having centre at $B$ and angle $\frac{B}{2}$.
Let $\rho_A$ a counter-clockwise rotation having center at $A$ and angle $\frac{A}{2}$.
To get $K$ we have to consider $\rho_B(I)$ then apply a dilation centered at $B$ with factor $\frac{BI}{B\rho_B(I)}$. Similarly, to get $M$ we have to consider $\rho_A(I)$ then apply a dilation centered at $A$ with factor $\frac{AI}{A\rho_A(I)}$. We have that both $\rho_B(IA)$ and $\rho_A(IB)$ are parallel to $AB$, so both $KI$ and $MI$ are parallel to $AB$ and the collinearity of $K,I,M$ is trivial.
A: Draw the triangle $BOA$ then from the condition we have that $\frac{BO}{AB} = \frac{BK}{BO}$. So using this choose $O'$ on $AB$ s.t. $BO = BO'$ and then draw a parallel to $AO$ that passes through $O'$. Denote the intersection of this line with $OB$ with $K'$. Then you have that $BK' = BK$ and you can find the point $K$. Now similarly you can construct $M$.
