$\triangle ABC$ is an equilateral triangle with circumcentre $O(1,2)$ and vertex $A$ lying on the line $5x+2y=4$.Find area of quadrilateral $BCDE$. $\triangle ABC$ is an equilateral triangle with circumcentre $O(1,2)$ and vertex $A$ lying on the line $5x+2y=4$. A circle with
centre $I(2,0)$ passes through the vertices $B$ and $C$ and intersects the sides $AC$ and $AB$ at $D$ and $E$ respectively. Find area of quadrilateral $BCDE$.
My Attempt
All I could do here was finding the perpendicular distance of $O$ from the given line. How do I use $I(2,0)$


As can be seen in figure OI is coincident on altitude of AH. The equation of AH is:
$y-2=\frac{-2}{2-1}(x-1)\Rightarrow y=-2x+4$
This with line $2y+6x=4$ gives $A(-4, 12)$ and we hav:
$AO=\sqrt {(1+5)^2+(12-2)^2}\approx 11.2$
$OH=\frac{AO}2=5.6$
$AH=\frac{3\times AO}2=16.8$
$AC=\frac{AH}{\sin 60}\approx 19.4$
$A_{ABC}=\frac{19.4\times 16.8}2=163$
$OI=\sqrt{1^2+5^2}=\sqrt 5\approx 2.24$
Now :
$IH=OH-OI=5.6-2.24=3.36$
$R=IC=\sqrt {3.36^2+(\frac{19.4}2)^2}\approx 10$
Equation of circle is:
$(x-2)^2+y^2=10^2$
This with line AH , $y=-2x+4$ gives:
$p(1-\sqrt 5, 2+2\sqrt 5$ and $Q(1+\sqrt 5, 2-2\sqrt 5)$
$AP=\sqrt{(-4-1+\sqrt 5)^2+(12-2-2\sqrt 5}\approx 6.2$
$ED=\frac{6.2}{16.8}\times 19.4\approx 7\Rightarrow A_{AED}=\frac 12(7\times 6.2)\approx 22$
$A_{BCDE}\approx 163-22=141$
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