Problem:
Triangle ABC has AB=13,BC=14, and AC=15. The points D,E, and F are the midpoints of AB,BC, and AC respectively. Let Xî€ =E be the intersection of the circumcircles of â–³BDE and â–³CEF. What is XA+XB+XC?
Answer Choices:
A. 24
B. 143​
C. 8195​
D. 141297​​
E. 4692​​ Solution:
Because DE is parallel to AC and EF is parallel to AB it follows that ∠BDE=∠BAC=∠EFC. By the Inscribed Angle Theorem, ∠BDE=∠BXE and ∠EFC=∠EXC. Therefore ∠BXE=∠EXC. Furthermore BE=EC, so by the Angle Bisector Theorem XB=XC. Note that ∠BXC=2∠BXE=2∠BDE=2∠BAC, and by the Inscribed Angle Theorem, it follows that X is the circumcenter of △ABC, so XA=XB=XC=R the circumradius of △ABC.
Let a=BC,b=AC, and c=AB. The area of △ABC equals 4R1​(abc), and by Heron's Formula it also equals s(s−a)(s−b)(s−c)​, where s=21​(a+b+c). Thus
