Chapter 15.3, Problem 14PSB

To determine

To Prove: The side BD is greater than FD in given triangle.

Explanation of Solution

Given:

  $\begin{array}{l}\to \Delta ACE\\ \to AC<AE\\ \to D\text{ is the midpoint of }\overline{CE}\\ \to \text{Point }B\text{ and }F\text{ are on }\overline{AC}\text{ and }\overline{AE}\\ \to \overline{CB}\cong \overline{FE}\end{array}$

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:

  

  $\begin{array}{l}\text{⇒}AC<AE\text{ , }\angle E<\angle C\\ \to \text{In }\Delta BCD\text{ and }\Delta DEF\\ \text{⇒}BC=EF\\ \text{⇒}DE=CD\\ \text{⇒}\angle E<\angle C\\ \text{⇒}\text{Conclusion : }BD>FD\left(\text{Hinge theorem}\right)\end{array}$

Conclusion:

Thus from calculation it is proved that $BD>FD$