Chapter 4.4, Problem 21PSC

To determine

To Prove: FC is perpendicular bisector of BD.

Explanation of Solution

Given information: F is midpoint of AE, ABCDE is regular pentagon.

Formula used: Properties of regular pentagon.

Proof:

  $\begin{array}{l}F\text{ is midpoint of }AE(\text{given})\\ \therefore AE=EF.\\ BC=CD(\text{given})\\ \therefore \angle CBD=\angle CDB\begin{array}{l}\text{(Angle opposites to equal sides}\hfill \\ \text{Of a triangle are equal)}\hfill \end{array}\\ BD\parallel AF\text{(ABCDE is regular pentagon)}\\ \therefore CF\text{bisects }BD(\therefore F\text{bisects }AE)\\ \therefore \text{Triangles formed by BC,CD and CF are congruent }\left(\text{SAS}\right)\\ \therefore \text{By linear pair and CPCT ,CF}\perp \text{BD}\text{.}\end{array}$