In triangle ABC, point X is on side BC such that AX=13,BX=10,CX=10, and the circumcircles of triangles ABX and ACX have the same radius. Find the area of triangle ABC.
The area of triangle ABC is
\(A_{ABC}=\dfrac{(\overline {BX}+\overline {XC})\cdot \overline {AX}}{2}=\dfrac{(10+10)\cdot 13}{2}=130\)
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