Question 704781: An isosceles triangle ABC has it vertices on a circle. If line(AB) =13cm, line(BC) =13cm and line(AC) = 10cm. Calculate the radius of the circle to the nearest whole cm.
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website!  CF is the altitude to base AB and part of AB's perpendicular bisector.
DE is part of BC's perpendicular bisector. (CDE is a right angle).
All points in a perpendicular bisector are equidistant from the ends of the segment.
Being on the perpendicular bisectos of AB and BC,
point E is at the same distance from A, B, and C,
so it is the center of the circle.
EC is the radius of the circle.
AB=10, AF=FB=10/2=5, BC=13, BD=CD=13/2=6.5
Using the Pythagorean theorem,
BCF and ECD are similar right triangles, because they have a right angle, and congruent angles (actually the same angle) at C.
The hypotenuse to long leg ratios on triangles BCF and ECD must be the same, because the triangles are similar triangles, so
 -->  -->  (rounded)
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