Chapter 15.3, Problem 26PSC

To determine

To Prove: The segment AD is greater than the BE in the given figure.

Explanation of Solution

Given:

  $\begin{array}{l}AC>CE\\ \overline{AB}\cong \overline{DE}\end{array}$

  

Formula Used:

The Hinge Theorem: It states that if two sides of two triangles are congruent and angle of one is greater than the angle of second then the third side of the triangle will be greater than the another.

Calculation:

  

  $\begin{array}{l}\to \text{In }\Delta ABC\\ \text{⇒}AC>CE\\ \text{⇒}\angle E>\angle A\\ \to \text{In }\Delta ADE\text{ and }\Delta ABE\\ \text{⇒}AE=AE\\ \text{⇒}AB=DE\\ \text{Conclusion:}\\ \text{⇒}\angle E>\angle A\\ \text{⇒}AD>BE\end{array}$

Conclusion:

Thusfrom the calculation it is proved that $AD>BE$