What If It's Zero? The other day I was deriving the combined equation for the angle bisectors of a pair of straight lines ($a{x^2} + 2hxy + b{y^2}$) passing through the origin. After some simplification, I got this equation - $$\left( {{m_1} + {m_2}} \right)\left( {{x^2} - {y^2}} \right) = 2xy\left( {1 - {m_1}{m_2}} \right)$$ 
where ${m_1}$ and ${m_2}$ are the slopes of the straight lines. By substituting ${m_1} + {m_2} = {{ - 2h} \over b}$ and ${m_1}{m_2} = {a \over b}$,$$\left( {{{ - 2h} \over b}} \right)\left( {{x^2} - {y^2}} \right) = 2xy\left( {{{b - a} \over b}} \right)$$
Since $b$ was a constant (coefficient of ${{y^2}}$), I multiplied both sides by $b$ and eliminated it.
But if $b$ happened to be zero, the equation would have been of the form $\infty  = \infty $, and by multiplying $0$ on both sides, the result would be indeterminate. 
How do the proofs tackle the case when denominator can also take the value zero? 
 A: Your problem came earlier; whatever method you used to determine $m_1 + m_2 = -2h/b$ and $m_1 m_2 = a/b$ is inapplicable in the case that $b=0$.
The usual fix to things like this is to split the problem into two parts:


*

*Solve the problem under the assumption that $b \neq 0$

*Solve the problem under the assumption that $b = 0$


And then combine the two solutions in whatever manner is appropriate to the problem.

As an example of this method, consider the problem of finding under what conditions the equation
$$ a x^2 + bx + c = 0$$
has a unique solution for $x$. We'd be tempted to use the quadratic formula, but that depends on $a$ being nonzero. So we split the problem:


*

*Assuming $a \neq 0$, we apply the quadratic formula; the solution is unique if and only if $b^2 - 4ac = 0$

*Assuming $a = 0$, we have a linear equation $bx + c = 0$. This has a unique solution if and only if $b \ne 0$


So, the end result is that the equation above has a unique solution for $x$ if and only if one of the following two conditions are satisfied:


*

*$a \neq 0$ and $b^2 = 4ac$

*$a = 0$ and $b \neq 0$

A: The short answer is that you have to treat those cases separately. Hurkyl’s answer covers this quite well. In particular, having to deal with $b=0$ separately comes up much earlier in your calculations.  
Sometimes having these special cases is unavoidable, but as I mentioned in my comment to your question, they can often be avoided in problems like this one if you use a different form of equation for the lines, as I’ll illustrate below.  
Let the two lines be given by the equations $Ax+By=0$, $Cx+Dy=0$, with $A$, $B$ not both zero and $C$, $D$ not both zero. The combined equation is $$(Ax+By)(Cx+Dy)=AC\,x^2+(AD+BC)\,xy+BD\,y^2=0.$$ Comparing this to your original equation, we must have $$\begin{align} a&=AC \\ b&=BD \\ h&=\frac12(AD+BC).\end{align}$$ The normal vectors to these lines are $\mathbf n_1=(A,B)$ and $\mathbf n_2=(C,D)$ with corresponding direction vectors $\mathbf v_1=(-B,A)$ and $\mathbf v_2=(-D,C)$. The two angle bisectors are then $$\mathbf w_1={\mathbf v_1\over\|\mathbf v_1\|}+{\mathbf v_2\over\|\mathbf v_2\|} \\ \mathbf w_2={\mathbf v_1\over\|\mathbf v_1\|}-{\mathbf v_2\over\|\mathbf v_2\|}.$$ These two vectors are orthogonal, so we can use them for the normals of the two bisecting lines. (It also means that we could’ve used the normals to the original lines directly to find the bisectors.) So, we have for the combined equation of the bisecting lines $(\mathbf w_1\cdot\mathbf x)(\mathbf w_2\cdot\mathbf x)=0$, which after expanding and a bit of fiddling becomes $${AD-BC\over(A^2+B^2)(C^2+D^2)}\left[(AD+BC)(x^2-y^2)-2(AC+BD)\,xy\right]=0.$$ By hypothesis, the denominator is non-zero. Since we have two distinct lines, $ab-h^2\lt0$, which after substitution is $\frac14(BC-AD)^2\gt0$, so in particular $AD-BC\ne0$. The above equation is therefore equivalent to $(AD+BC)(x^2-y^2)-2(AC+BD)\,xy=0$, and substituting the original coefficients into this results in $$h(x^2-y^2)-(a-b)xy=0.$$ As you can see, by starting with the point-normal form of the equations instead of the slope-intercept form, we never had to deal with vertical lines (or any resulting vertical bisectors) separately.
