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Let angle ABC be called B and angle ACB be called C. Let angle PBC be called B' and angle QCB be called C'.

B' < C' (B' is less than C') ----(1) B' = 180 - B, C = 180 - C, Substituting in (1), 180 - B < 180 - C OR - B < - C OR B > C (B is greater than C)

In a triangle, side opposite the greater angle is greater than the side opposite the smaller angle, Hence side AC (opposite B, greater angel) > side AB (opposite C,smaller angle)

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See the diagram.

Angle PBC < angle QCB So, x < y So, 180 - x > 180 - y

Angle ABC > angle ACB So, angle B > angle C

Now we will prove that AC > AB.

Draw a line BD from B such that the angle ABD is equal to angle C. Draw a line BE from B such that the angle ABE is equal to (B+C)/2.

Angle ABE = (B+C)/2 and angle AEB = 180 - A - angle ABE = 180 - A - (B+C)/2 = 180 - (180- B - C) - (B+C)/2 = (B+C)/2 = angle ABE

So , ABE is an isosceles triangle. Sides AB = AE. Now,