Question

Sides AB and AC of ΔABC are extended to pts P and Q respectively. Also angle PBC is less than angle QCB. Show that AC is greater than AB.

SUPER URGENT!!!! PLEASE!!!!!

Answer 1

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) 
 

Answer 2

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,

AC = AE + EC 

AC - AB = AE + EC - AB = EC  > 0

So AC > AB