Chapter 5.5, Problem 23PPS

To determine

To prove: The statement of inequality $AC+BC>AB$ from the given figure.

Explanation of Solution

Given information:

The given diagram is:

  

Proof:

Consider the given diagram of the triangle:

  

Draw an angle bisector of angle A on BD

  

From the definition of congruency, we can write that,

  $CD=AC$

From the definition of isosceles triangle theorem, we can write that,

  $\begin{array}{l}\angle CAD\cong \angle CDA\\ m\angle CAD\cong m\angle CDA\end{array}$

From the above diagram, we can write:

  $m\angle BAC+m\angle CAD\cong m\angle BAD$

Substitute $\angle CDA$ for $\angle CAD$ .

  $m\angle BAC+m\angle CDA\cong m\angle BAD$

Now, consider the inequality of triangle theorem:

  $m\angle CDA<m\angle BAD$

Consider the angle side relationship of triangles

  $AB<BD$

From the above diagram write:

  $BD=CD+BC$

Substitute AC for CD as $CD=AC$ .

  $BD=AC+BC$

Substitute AB for BD as $AB<BD$ .

  $AB<AC+BC$

Hence, the statement is proved.