ABC is a equilateral triangle .
Proof :-
Let's assume both circles as C1 and C2 [ as shown in the figure ]
- AB is the radius of circle C1
- AB is the radius of Circle C2
AC is the radius of circle C1.
BC is the radius of circle C2 .
AB and AC both are radius of circle C1 so both are equal ie AB = AC .
AB and BC both are radius of circle C 2 so both are equal ie AB = BC .
Hence we conclude that .
AB = BC = AC.
So the triangle is equilateral triangle.
This should be recognized as the difference of perfect squares which is of the form:
(a^2-b^2) and the difference of squares always factors to:
(a-b)(a+b) in this case:
(3x-8)(3x+8)