A circle meets the sides of an equilateral triangle ABC at six points D, E, F ,G, H , I in the figure . If AE= 4 ED = 26 , FG = 14 , and the circle with diameter HI has area πb, find b.
sorry i don't how draw a picture and how to use mathematics symbols here..this is reason i added a picture.. thanks all of u to answer my questions.
The length of the side of the equilateral triangle is $4+26+2=32$.
By this, we have
$$AE\times AD=AF\times AG$$
This implies $$AF=6,BG=12.$$
Also, letting $x=HI, y=BH$, we have$$BG\times BF=BH\times BI\iff 12\times 26=y(x+y)$$ $$CD\times CE=CI\times CH\iff 2\times 28=(32-x-y)(32-y)$$
Solving these for $x$ gives us $x=HI=4\sqrt{22}$. Hence, we have $$b=\frac{x^2}{4}=88.$$