We have two lines :
$$L_1 : a_1x+b_1y+c_1=0,\quad L_2 : a_2x+b_2y+c_2=0$$
and the angle bisectors :
$$L_{\pm} : \frac{a_1x+b_1y+c_1}{\sqrt{a_1^2+b_1^2}}=\pm\frac{a_2x+b_2y+c_2}{\sqrt{a_2^2+b_2^2}}$$
If we let $\theta$ be the (smaller) angle between $L_+$ and $L_1$, then we have
$$\cos\theta=\frac{\left|a_1\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)+b_1\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)\right|}{\sqrt{a_1^2+b_1^2}\sqrt{\left(\frac{a_1}{\sqrt{a_1^2+b_1^2}}-\frac{a_2}{\sqrt{a_2^2+b_2^2}}\right)^2+\left(\frac{b_1}{\sqrt{a_1^2+b_1^2}}-\frac{b_2}{\sqrt{a_2^2+b_2^2}}\right)^2}}$$
$$=\frac{\left|\sqrt{a_1^2+b_1^2}-\frac{a_1a_2+b_1b_2}{\sqrt{a_2^2+b_2^2}}\right|}{\sqrt{a_1^2+b_1^2}\sqrt{2-2\frac{a_1a_2+b_1b_2}{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}}}\times\frac{2\frac{1}{\sqrt{a_1^2+b_1^2}}}{2\frac{1}{\sqrt{a_1^2+b_1^2}}}=\sqrt{\frac{1-\frac{a_1a_2+b_1b_2}{\sqrt{(a_1^2+b_1^2)(a_2^2+b_2^2)}}}{2}}$$
Hence, we can see that
$$\begin{align}a_1a_2+b_1b_2\gt 0&\iff\cos\theta\lt 1/\sqrt 2\\&\iff \theta\gt 45^\circ\\&\iff \text{$L_+$ is the obtuse angle bisector}\end{align}$$
as desired.
(Note that "$c_1,c_2$ both are of same sign" is irrelevant.)
The formula
$$ d_1(x,y) = \frac{\lvert a_1x+b_1y+c_1 \rvert}{\sqrt{a_1^2+b_1^2}} $$
gives you the distance from a point $(x,y)$ to the line $L_1$ whose equation
is $a_1x+b_1y+c_1 = 0.$
See
Distance Between A Point And A Line
for a proof of this.
For the line $L_2$ given by $a_2x+b_2y+c_2 = 0,$ the distance of a point to the line is given by
$$ d_2(x,y) = \frac{\lvert a_2x+b_2y+c_2 \rvert}{\sqrt{a_2^2+b_2^2}}. $$
A point $(x,y)$ on an angle bisector between two lines is equidistant from the two lines, that is, it satisfies the condition
$d_1(x,y) = d_2(x,y).$ Writing out the formulas for $d_1$ and $d_2$ in full,
$$ \frac{\lvert a_1x+b_1y+c_1 \rvert}{\sqrt{a_1^2+b_1^2}}
= \frac{\lvert a_2x+b_2y+c_2 \rvert}{\sqrt{a_2^2+b_2^2}}. $$
Now observe that $\lvert a_1x+b_1y+c_1 \rvert$ will be either
$a_1x+b_1y+c_1$ or $-(a_1x+b_1y+c_1),$ whichever of those two
expressions is positive.
In fact, $a_1x+b_1y+c_1$ will be positive for all points on one side
of the line and negative for all points on the other side.
Now if the lines $L_1$ and $L_2$ intersect,
they divide the plane into four regions.
Label each these regions as $+L_1$ or $-L_1$ depending on whether
$a_1x+b_1y+c_1$ is (respectively) positive or negative in that region.
Label each region as $+L_2$ or $-L_2$ depending on whether
$a_2x+b_2y+c_2$ is (respectively) positive or negative in that region.
One of the angle bisectors of $L_1$ and $L_2$ will go through the regions
labeled $+L_1,+L_2$ or $-L_1,-L_2.$
That is, on that line the signs of $a_1x+b_1y+c_1$ and $a_2x+b_2y+c_2$
are either both positive or both negative.
Points on this line therefore satisfy the formula
$$ \frac{a_1x+b_1y+c_1 }{\sqrt{a_1^2+b_1^2}}
= \frac{a_2x+b_2y+c_2 }{\sqrt{a_2^2+b_2^2}}. $$
(For points in the region $-L_1,-L_2,$ this formula gives negative values
on both sides, but their absolute values are equal.)
The other angle bisector goes through
$+L_1,-L_2$ and $-L_1,+L_2$ and has the formula
$$ \frac{a_1x+b_1y+c_1 }{\sqrt{a_1^2+b_1^2}}
= - \frac{a_2x+b_2y+c_2 }{\sqrt{a_2^2+b_2^2}}. $$
Best Answer
In general, the equation $\ ax^2+2hxy+by^2+2gx+2fy+c=0\ $ defines a conic, of which two intersecting straight lines is one (degenerate) special case. One standard way to determine what form of conic the equation represents is to diagonalise the matrix, $$ A=\pmatrix{a&h\\h&b}\ $$ by finding its eigenvalues and egenvectors. Let $\ \lambda_1, \lambda_2\ $ be the eigenvalues, and $\ \boldsymbol{e}_1, \boldsymbol{e}_2\ $ the corresponding normalised eigenvectors, which can be chosen so that $\ \boldsymbol{e}_1=\pmatrix{\cos\theta\\-\sin\theta}\ $ and $\ \boldsymbol{e}_2=\pmatrix{\sin\theta\\\cos\theta}\ $ for some angle $\ \theta\ $. If $\ \Theta\ $ is the matrix with columns $\ \boldsymbol{e}_1\ $ and $\ \boldsymbol{e}_2\ $, then $\ \Theta\ $ is a rotation matrix, with $\ \Theta^{-1} = \Theta^\top\ $, and $$ \Theta^\top A\Theta = \pmatrix{\lambda_1 & 0\\0&\lambda_2}\ . $$ Now, the equation we are interested in can be written as \begin{eqnarray} 0&=& \boldsymbol{x}^\top A\boldsymbol{x} + 2\boldsymbol{g}^\top\boldsymbol{x} + c\\ &=& \left(\Theta^\top \boldsymbol{x}\right)^\top\Theta^\top A\Theta\left(\Theta^\top\boldsymbol{x}\right) + 2\left(\Theta^\top\boldsymbol{g}\right)^\top \Theta^\top\boldsymbol{x}+c\\ &=&\boldsymbol{x}'^\top\Lambda\boldsymbol{x}'+2\boldsymbol{g}'^\top\boldsymbol{x}'+c\ , \end{eqnarray} where $\ \boldsymbol{x}=\pmatrix{x\\y}\ $, $\ \boldsymbol{x}'=\Theta^\top\boldsymbol{x}\\$, $\ \Lambda = \pmatrix{\lambda_1 & 0\\0&\lambda_2}\ $, $\ \boldsymbol{g}=\pmatrix{g\\h}\ $, and $\ \boldsymbol{g}'=\Theta^\top\boldsymbol{g}\ $. The entries of $\ \boldsymbol{x}'\ $, $\ x_1'=x\cos\theta+y\sin\theta\ $, and $\ x_2'=-x\sin\theta+y\cos\theta\ $, are the coordinates of a point $P$ with respect to a set of axes that have been rotated clockwise through an angle $\ \theta\ $relative to the original axes, where $\ x\ $ and $ y\ $ are the coordinates of $P$ with respect to those original axes. It follows from above that the equation the the conic with respect to the new axes is \begin{eqnarray} 0 &=& \lambda_1 x_1'^2 + \lambda_2x_2'^2 +2g_1'x_1'+ 2g_2'x_2' + c\\ &=& \lambda_1\left(x_1' +\frac{g_1'}{\lambda_1}\right)^2 + \lambda_2\left(x_2' +\frac{g_2'}{\lambda_2}\right)^2 +c - \frac{g_1'^2}{\lambda_1}-\frac{g_2'^2}{\lambda_2}\ . \end{eqnarray} This is the equation of two intersecting straight lines if and only if $\ \lambda_1\ne0\ $, $\ \lambda_2\ne0\ $, $\ \lambda_1\ $ and $\ \lambda_2\ $ are of opposite sign, and $\ c - \frac{g_1'^2}{\lambda_1}-\frac{g_2'^2}{\lambda_2}=0\ $. If this is the case, suppose, without loss of generality, that $\ \lambda_1>0\ $ and $\ \lambda_2<0\ $. Then the above equation becomes \begin{eqnarray} 0 &=& \left(\sqrt{\lambda_1}x_1' + \frac{g_1'}{\sqrt{\lambda_1}}\right)^2- \left(\sqrt{-\lambda_2}x_2' - \frac{g_2'}{\sqrt{-\lambda_2}}\right)^2\\ &=& \left(\sqrt{\lambda_1}x_1' + \sqrt{-\lambda_2}x_2'+ \frac{g_1'}{\sqrt{\lambda_1}}-\frac{g_2'}{\sqrt{-\lambda_2}}\right)\\ &&\ \ \ \cdot \left(\sqrt{\lambda_1}x_1' - \sqrt{-\lambda_2}x_2'+ \frac{g_1'}{\sqrt{\lambda_1}}+\frac{g_2'}{\sqrt{-\lambda_2}}\right)\ , \end{eqnarray} and the equations of the two straight lines in the new coordinates are \begin{eqnarray} \sqrt{\lambda_1}x_1' + \sqrt{-\lambda_2}x_2'+ \frac{g_1'}{\sqrt{\lambda_1}}-\frac{g_2'}{\sqrt{-\lambda_2}}&=&0\ \ \mbox{, and}\\ \sqrt{\lambda_1}x_1' - \sqrt{-\lambda_2}x_2'+ \frac{g_1'}{\sqrt{\lambda_1}}+\frac{g_2'}{\sqrt{-\lambda_2}} &=&0\ . \end{eqnarray} Finally, substituting $\ x_1'=x\cos\theta+y\sin\theta\ $, and $\ x_2'=-x\sin\theta+y\cos\theta\ $ in these equations, we get the equations of the lines in the original coordinates: \begin{eqnarray} x\left(\sqrt{\lambda_1}\cos\theta - \sqrt{-\lambda_2}\sin\theta\right)&+ &y\left(\sqrt{\lambda_1}\sin\theta +\sqrt{-\lambda_2}\cos\theta\right)\\ &+& \frac{g_1'}{\sqrt{\lambda_1}}-\frac{g_2'}{\sqrt{-\lambda_2}}&=&0 \end{eqnarray} and \begin{eqnarray} x\left(\sqrt{\lambda_1}\cos\theta + \sqrt{-\lambda_2}\sin\theta\right)&+ &y\left(\sqrt{\lambda_1}\sin\theta -\sqrt{-\lambda_2}\cos\theta\right)\\ &+& \frac{g_1'}{\sqrt{\lambda_1}}+\frac{g_2'}{\sqrt{-\lambda_2}}&=&0\ . \end{eqnarray}