Proving a Line to bisect a line in a Triangle From point $A$ tangents $AB$ and $AC$ to a circle are drawn ($B$ and $C$ tangent points); $PQ$ is a diameter of the circle; line $L$ is tangent to the circle at point $Q$. Lines $PA$, $PB$, and $PC$ intersect line $L$ at points $A_1, B_1, C_1$. Prove that $A_1B_1 = A_1C_1$. 

Not a complete solution yet, but this is the beginning of pure "bashing" with coordinates. 
Choose a coordinate system so that $B_1 = (1,0)$ and $C_1 = (0, -1).$ Then, it suffices to prove that $A_1 = (0,0).$
Let $P = (a,2b)$ and $Q = (a,0)$, so the equation of the circle is $(x-a)^2+(y-b)^2 = b^2$. This is simplified to $x^2+y^2-2ax-2by+a^2 = 0.$ The equation of the line passing through the points $P$ and $B_1$ is simply $y=\dfrac{2b}{a-1}x-\dfrac{2b}{a-1}$. Thus, coordinate of the point $B$ will satisfy the above two equations, since it is the intersection of the circle and that line. 
Substituting $y$ in terms of $x$ and denote $p = \dfrac{2b}{a-1}$, we get: $$0 = (1+p^2)x^2-2x(a+bp+p^2)+(p^2+2bp+a^2).$$
Now when you solve this equation, you will get two solutions $x_1 = a$ and $x_2 = \dfrac{p^2+a}{p^2+1}$, but we have to take the latter because $P\neq B.$ Thus, $B = \Big(\dfrac{p^2+a}{p^2+1}, \dfrac{2b}{p^2+1}\Big)$. 
Similarly, the equation of the line passing through $P$ and $C$ is $y = \dfrac{2b+1}{a}x - 1$ and when you plug this into the equation of the circle you will find $C = \Big(\dfrac{q+a}{q^2+1}, \dfrac{aq-1}{q^2+1} \Big)$, where $q = \dfrac{2b+1}{a}.$
Now, it is straightforward to find the coordinate of point $A$ because it is the intersection of two tangents to the circle at $B$ and $C$. After this, find the equation of the line $l = PA$, then the coordinate of $A_1$ will be found by intersecting $l$ with the $x$ axis. 
I will not have time to complete the remaining computations, but it is completely straightforward at this point. But our goal is to prove that $A_1 = (0,0)$; therefore, if the problem is true, then it the equation of the line $l$ must have been found in the form $y = cx, c\in\mathbb{R}.$
