Prove that:

1. In a triangle angles opp. to longer side is larger.

2. Also, prove its converse.

4)proof  C

  D

  A  B

Let ABC be a triangle in which the side AC is longer than the side BC

We need to prove that angle ABC is greater than angle BAC

On the ray CA representing the longer side , mark the segment CD equal to the smaller side BC. 

Since the side BC is shorter than AC , the point D is located between A and C

Draw the segment , BD joining point B and D . this segment is located inside the triangle , therefore angle ABC is greater than DBC

On the other hand .

Angle DBC = angle BDC  …………….(1)

 We know , angle ABC = Angle DBC + angle DBA

angle ABC = Angle BDC + angle DBA

also we know, angle BDC is exterior angle of triangle ADB

therefore angle ADB = angle DAB+ angle ABD

This mean angle BDC is greater than CAB.

angle ABC is greater than the angle DBC; the angle DBC is equal to the angle BDC, and the angle BDC is greater than the angle BAC

Therefore, angle ABC is greater than angle BAC

HENCE PROVED

CONVERSE:

Let ABC be a triangle in which the angle ABC is greater than the angle BAC  
(Figure 3). We need to prove that the side AC is longer than the side BC. 

Let us suppose that the opposite is true: the side AC is not longer than 
the side BC. Then two cases can occur: either AC is of the same length 
as BC or AC is shorter than BC. 

In the first case the triangle ABC would has been isosceles and the angles ABC and BAC would have been congruent (see the lesson 
This contradicts to the given condition

In the second case the side BC would has been longer than AC and, 
consequently, the angle BAC would has been greater than the angle ABC 
in accordance to the Theorem 1 proved in this lesson above. 
This contradicts to the given condition again.

Thus the only remaining possibility is that the side AC is longer than the side BC. This is exactly what to be demonstrated.