How can I prove the triangle is equilateral if the two circles have same diameter?
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Just show that triangles $AOB, BOC, COA$ are all congruent (then all the sides of triangle $ABC$ must be equal).
To show this, note that triangles $A'BO, A'OC$ are equilateral since they are all radius of the circles. Line $OA'$ is perpendicular to line $BC$. Hence, $\angle OBC = \angle OCB = 30^\circ$. So $\angle BOA' = \angle COA' = 60^\circ$. Now, $\angle BOA = 180^\circ - \angle BOA' = 180^\circ - 60^\circ = 120^\circ$ since $A'A$ is a straight line. Similarly, $\angle AOC = 120^\circ$. So $\angle BOC = \angle BOA = \angle AOC = 120^\circ$. Now its easy to show triangles $AOB, BOC, COA$ are all congruent since they are isosceles and have common equal sides.
Hint: join the two dots (the points where the horizontal line intersects the two circles) to the point just above them where the two circles intersect. What can you say about the triangle you have just formed?
Call the rightmost vertex of the triangle $\;A\;$ , the upper one $\;B\;$ and the lower one $\;C\;$, let $\;M,N\;$ be the left (right) circle's center and let $\;P\;$ be the intersection point of $\;AM\;$ with $\;BC\;$. Be sure you can prove (or at least follow) the following:
=== $\;MN\perp BC\;$ (the center's segment is always perpendicular to the common cord of two intersecting, non-tangent, circles)
=== Thus, $\;BP=PC\;$ (a straight segment through a circle's center is perpendicular to a cord iff it bisects it)
=== in $\;\Delta ABC\;$ , we have that $\;AP\perp BC\;$ is also the median to $\;BC\;$ and thus $\;\Delta ABC\;$ is isosceles, with $\;AB=AC\;$
=== $\;BMCN\;$ is a rombus, and thus $\;BM=MC=$radius
=== $\;\Delta NBM\;$ is equilateral and since $\;\angle BNM\;$ is a central angle in the right circle and it equals... then...finish the exercise.