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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Theorem 8.11 in what book?
I will give a link that explains and proves the theorem.
in this theorem we will neglect the given resistance and in next step mean as second step we will solve
Well, this will depend on the length of the sides of the triangle for what postulate or theorem you will be using.
SAS