Prove that a side of a triangle is greater than its semiperimeter. This is one of the question in my text book and I dont think I understood this correctly. If we think of a triangle whose all sides are 5 cm. Then, the semiperimeter of this triangle would be 7.5 cm which is greater than a side 5 cm. Since the side is less than the semiperimeter, either I understood the question wrong or the question has mistake. Any comment?
 A: Given a triangle with side A. 
Draw side B from one vertex of A. 
Draw side C from the other vertex of A. 
If B+C were less than the length of A, then sides B and C would not meet. 
If side B+C are equal to A then the no triangle would be possible ie no enclosed space. 
So B+C > A The side of a triangle (A) is less than the sum of the other 2 sides (B+C). 
A+(B+C)>A+A  Adding A to both sides does not change the inequality
(A+B+C)/2 > 2*A/2 Dividing by 2 on both sides does not change the inequality. 
The semiperimeter > A   Definition of semiperimeter and algebraic simplification. 
The semiperimeter of a triangle is greater than the length of an arbitrarily chosen side A. 
Note if draw line segments equal to B and C above line A, you should notice that the semiperimeter is equal to side A...plus half of (B+C)-A. 
This makes the semiperimeter an average between a selected side....and the sum of the other 2 sides. 
Keselev Geometry Plainemetry page 41 question 89 states 
Prove theorems: A side of a triangle is GREATER than its semiperimeter. 
This is an ERROR in the textbook. One side of a triangle is ALWAYS less than its semiperimeter. Please change the word Greater to LESS for this question. 
