Chapter 5.5, Problem 18PPS

To determine

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

Explanation of Solution

Given information:

The given congruency statement is:

  $\angle BCD\cong \angle CDB$

The given diagram is:

  

Consider the given statement of inequality as $\angle BCD\cong \angle CDB$ .

Recall the converse of the isosceles triangle theorem as:

  $BC\cong BD$

From the definition of congruency, we can write that,

  $BC=BD$

Now, consider the inequality of triangle theorem:

The sum of the lengths of any two sides of a triangle should be greater than the third side.

  $AB+AD>BD$

Consider the third side of the triangle as $BC$ .

Since, $BC=BD$

  $AB+AD>BC$

Hence, the statement is proved.