Find semi perimeter and Area of triangle in terms of $d_1,d_2,d_3,R$ and $|D|$. 
For $i=1,2,3$ ; Let $L_i=\{(x,y):x\cos\theta_i+y\sin\theta_i=1\}$ , $A,B$ and $C$ are intersections of $(L_2,L_3),(L_3,L_1)$ and $(L_1,L_2)$ respectively. And $d_1,d_2$ and $d_3$ are lengths of altitudes of the triangle from  $A,B$ and $C$, respectively. It is given that $A,B$ and $C$ defined above are real points and
D=$\begin{bmatrix} \cos\theta_1 & \sin\theta_1& 1 \\ \cos\theta_2& \sin\theta_2& 1 \\\cos\theta_3& \sin\theta_3&  1  \end{bmatrix}$ , $|D|\ne0$ , R is the circumradius of triangle ABC
. Find the semi perimeter and Area of triangle in terms of $d_1,d_2,d_3,R$ and $|D|$.

My attempt :
I calculated the co-ordinates of $A,B$ and $C$ , and side lengths of the triangle in terms of $\theta_i$ , and also calculated value of $|D|$ , then I got the desired result after plugging in the gigantic values . But I noticed that D is the matrix formed by coefficients of lines $L_i$ so i think there might be some geometric significance to that matrix , and my intuition says that there must be a substantially less hectic way to solve it with the help of that , because the way I have solved it is too lengthy .
So could someone please tell me a better way of solving this question , that'll be a great help.
Thanks !
 A: Let $(x_i,y_i)$ (vertices of $ABC$) be the intersection of $L_j$’s with $j\neq i$ for $i=1,2,3$. The area of the triangle is half of the absolute value of the determinant $$\left|\begin{array}{ccc}x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1\end{array}\right|.$$
Now the key observation is that $$\left[\begin{array}{ccc}x_1&y_1&1\\ x_2&y_2&1\\ x_3&y_3&1\end{array}\right]\left[\begin{array}{ccc}\cos \theta_1&\cos\theta_2&\cos\theta_3\\ \sin\theta_1&\sin\theta_2&\sin\theta_3\\ -1&-1&-1\end{array}\right|={\rm diag}(\pm d_1,\pm d_2,\pm d_3),\quad (1)$$ where the last matrix is diagonal. This follows from the distance formula from $(x_i,y_i)$ to $L_j$, namely $$d((x_i,y_i),L_j)=\frac{|x_i\cos\theta_j+\sin\theta_j-1|}{\sqrt{\cos^2\theta_i+\sin^2\theta_i}}=\delta_{ij}d_i.$$ The second important observation is that $$d((0,0),L_i)=1$$ showing that $(0,0)$ is the incenter of triangle $ABC$ with inradius $1$. Now the rest is standard. Let’s denote the area of $ABC$ by $\Delta$. Then from (1), one has $$2\Delta\cdot|D|=d_1d_2d_3,$$ so $$\Delta=\frac{d_1d_2d_3}{2|D|}.$$ Then use the fact that $$\Delta=\frac{abc}{4R}=sr$$ and $$a=\frac{2\Delta}{d_1},b=\frac{2\Delta}{d_2},c=\frac{2\Delta}{d_3}$$ to finish up, where $R$ (resp. $r=1$) is the circumradius (resp. inradius) of $ABC$, $a,b,c,$ the side lengths and $s$ is the semiperimeter of $ABC$.
