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Stewart's Theorem

​
fraction of distance from B to C
0.3
BC
×
2
AP
≈
811.28
2
AC
×
BP
+
2
AB
×
CP
-
BC
×
BP
×
CP
≈
811.28
Stewart's theorem states that if in a triangle ABC a line is drawn from A to a point P on BC, then
BC
2
AP
=
2
AC
BP+
2
AB
CP-BCBPCP
.
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