![I created this figure using Excel, then copied it to Irfanview]()
With reference to the figure above:
Considering the triangle ABD:
(Note: since the triangle ABC is equilateral, its internal angles are equal to 60^@60∘and a circle's radius with an endpoint in the triangle vertex A bisects the internal angle B hat A CBˆAC in two angles, B hat A DBˆAD and D hat A CDˆAC, of 30^@30∘ each.)
tan B hat A D=tan 30^@=("side"/2)/(height)tanBˆAD=tan30∘=side2height => height = "side"/(2*(sqrt(3)/3))=sqrt(3)/2*"side"height=side2⋅(√33)=√32⋅side
S_(triangle)=(("base")("height"))/2=(("side")(sqrt(3)/2)("side"))/2=(sqrt(3)/4)("side")^2S△=(base)(height)2=(side)(√32)(side)2=(√34)(side)2 => (cancel(sqrt(3))/4)("side")^2=4*cancel(sqrt(3)) => ("side")^2=16 => side=4
Considering the triangle BDE:
(Note that all the 3 triangles resulting from the division of the triangle ABC by the circle's radii with endpoints in A, B and C (triangles ABE, BCE and ACE) are congruent. So since their obtuse angles are equal and sum 360^@ (A hat E B+B hat E C+A hat E C=360^@), each one of them must be equal to 360^@/3=120^@. Since the triangles BDE and CDE are congruents, B hat E D=C hatE D, and since they sum 120^@ (=B hat E C), each one of them is equal to 60^@).
sin B hat E D= sin 60^@=("side"/2)/R => R="side"/(2*sin 60^@)=(4)/(cancel(2)*sqrt(3)/cancel(2))=4/sqrt(3)
Circle's Area
S_(circ)=pi*R^2=pi*(4/sqrt(3))^2=(16/3)pi
