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If of triangle ABC and A'B'C' sides AB = A'B' and AC = A'C', and the included angle at A is larger than the included angle at A*, then BC > B'C'.
Proof:
A A'
/|\ /|
/ | \ / |
/ | \ / |
/ | \ B'/ |
B | X \C |C'
D
We construct AD such that AD = A'C' = AC and angle BAD = angle B'A'C'.
Triangles ABD and A'B'C' are congruent. Therefore BD = B'C'.
Let X be the point where the angle bisector of angle DAC meets BC.
From the congruent triangles AXC and AXD (SAS) we have that XD = XC.
Now, by the triangle inequality we have that BX + XD > BD, so BX + XC > BD, and consequently BC > BD = B'C'.
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