Pair of straight lines in 2D A triangle has the lines $ax^2+2hxy+by^2=0$ for two of its sides and the point $(c,d)$ for its orthocenter. Prove that the equation of the third line is 
$(a+b)(cx+dy)=ad^2-2hcd+bc^2$
My approach : I found the slope of the line as $-c/d$. To get the coordinates of a point on the line I assumed it to be point $(x1,y1)$ which is intersection of this line with one of the other pair of lines. But elimination is very tedious. Please suggest aliter or how to do elimination. 
 A: There’s a somewhat obscure but fairly easy to prove geometric fact that makes the calculations for this problem much simpler: if you construct a circle with one of the sides of the triangle as a diameter, the feet of the altitudes from the vertices on this side lie on the circle. Another way to put this is that if two vertices and the foot of an altitude from either vertex lie on a circle, then those two vertices are the ends of a diameter.
 
If the equation of the extensions of two of the triangle’s sides is $ax^2+2hxy+by^2=0$, then we know that one of the vertices is the origin, and that an equation of the perpendiculars to these sides through the point $H(c,d)$ is $b(x-c)^2-2h(x-c)(y-d)+a(y-d)^2=0$. Adding these two equations produces $$(a+b)(x^2+y^2)-2(bc-dh)x-2(ad-ch)y+ad^2-2hcd+bc^2=0.$$ This is the equation of a circle that passes through the four points of intersection of the previous two pairs of lines, namely, the two unknown vertices of the triangle and the feet of the altitudes from those vertices. The center of this circle, which we can read from its equation, is at $x={bc-dh\over a+b}$, $y={ad-ch\over a+b}$ and is the midpoint of the missing side. This side is perpendicular to the altitude from the origin, which goes through $H$, so an equation of its extension has the form $$cx+dy=cx_0+dy_0,$$ where $(x_0,y_0)$ is any point on the line. Substituting the coordinates of the center of the circle and rearranging, we obtain $$(a+b)(cx+dy)=ad^2-2hcd+bc^2,$$ as required.  
More valuable than the above derivation might be the way that I arrived at it, which was basically by reverse-engineering the given equation of the third side. I recognized the right-hand side as the constant term of the equation of the perpendiculars through $H$, which suggested playing around with that equation. It certainly could be relevant since it covers two of the three altitudes. Adding it to the equation of the known sides eliminated the cross terms and produced a common factor of $a+b$ for the remaining squared terms, which looked promising. The rest was a matter of working out the relationship between the resulting circle and the triangle, which became pretty obvious after drawing a couple of concrete examples.
A: 
From the given equation of the pair of straight lines, it is evident that they both pass through the origin of the coordinate system (see $\mathrm{Fig. 1}$). If we denote the two sides lying on these lines $AC$ and $AB$, their respective slopes can be easily determined as,
$$AC:\space \frac{-h+\sqrt{h^2-ab}}{b}\qquad \mathrm{and}\qquad AB:\space \frac{-h-\sqrt{h^2-ab}}{b}. \tag{1}$$
It is obvious that their point of intersection, which serves as one of the vertices of the triangle, is $A$. We also let the vertex $C: \left(x_1,\space y_1\right)$. Only those lines that have the color red are considered in our proof. 
What we seek is the equation of the side $BC$. The altitude $AD$ that is perpendicular to this side passes through the orthocenter $H$ and the origin $A$. Therefore, the slope of $AD$ is $\left(\cfrac{d}{c}\right)$. Then, we know that the slope of $BC$ is equal to 
$\left(-\cfrac{c}{d}\right)$.
So, that’s where you got yourself bogged down. You probably felt little lazy, didn't you?. According to my opinion, you should have continued and carried out the steps you have proposed, because they are spot-on. 
The equation of $BC$ can be written as
$$cx+dy=k, \tag{2}$$
where $k$ is an unknown. Since $C$ lies on the side $BC$, we have
$$cx_1+dy_1=k \tag{3}.$$
Our aim is to express $x_1$ and $y_1$ in terms of $a$, $b$, $c$, $d$, and $h$, so that we can substitute them in equation (3) to obtain a value for $k$. 
Consider the side $AC$ and the altitude $CF$. In each case, we know the slope and a point on the line. Therefore, the respective equations of the lines $AC$ and $CF$ can be write down as,
$$
    \begin{matrix}
   \qquad\qquad\qquad\qquad\qquad\qquad & y & = & \cfrac{-h+\sqrt{h^2-ab}}{b}x,\qquad\qquad & \qquad\qquad\qquad\qquad (4) \\
   \qquad\qquad\qquad\qquad\qquad & y & = & \cfrac{b}{h+\sqrt{h^2-ab}}\left(x-c\right)+d.  & \qquad\qquad\qquad\qquad (5)  \\
    \end{matrix}
$$
Since the point $C$ is on both $AC$ and $CF$, we shall have,
$$y_1=\frac{-h+\sqrt{h^2-ab}}{b}x_1 =\frac{b}{h+\sqrt{h^2-ab}}\left(x_1-c\right)+d, \tag{6}$$
which allows us to find a value for $x_1$, i.e.
$$x_1=\frac{bc-d\left(h+\sqrt{h^2-ab}\right)}{a+b}. \tag{7}$$
Now, to find $y_1$, we need to substitute the value of $x_1$ from equation (7) in equation(6), i.e.
$$y_1=\frac{ad+c\left(-h+\sqrt{h^2-ab}\right)}{a+b}. \tag{8}$$
Substitution of values obtained in equations (7) and (8) in equation (3) yields 
$$k= \frac{bc^2-cd\left(h+\sqrt{h^2-ab}\right)}{a+b}+\frac{ad^2+cd\left(-h+\sqrt{h^2-ab}\right)}{a+b}.$$
Simplification of the above expression leads to
$$k=\frac{bc^2-2cdh+ad^2}{a+b}.$$
Finally we substitute this value of $k$ in equation (2) to obtain the equation of the side $BC$, which is shown below.
$$\left(a+b\right)\left(cx+dy\right)=bc^2-2cdh+ad^2.$$
