Chapter 5.3, Problem 25WE

To determine

To prove:  $DE=EF$

Explanation of Solution

Given information: 

  $\begin{array}{l}AB=BC\\ ACandDF\text{ are transversal line}\end{array}$

Proof:

  $\begin{array}{l}\text{In }\Delta \text{ABX and }\Delta \text{ACF}\\ \angle \text{ABX}=\angle ACF(BX||CF)\\ \angle AXB=\angle AFC(BX||CF)\\ \angle A=\angle A(common)\\ by\text{ AAA rule}\\ \Delta \text{ABX~}\Delta \text{ACF}\\ \text{so,}\\ \frac{AB}{AC}=\frac{BX}{CF}=\frac{AX}{AF}\end{array}$

So,

  $\begin{array}{l}\frac{AB}{AB+BC}=\frac{AX}{AX+XF}\\ \frac{AX+XF}{AX}=\frac{AB+BC}{AB}\\ 1+\frac{XF}{AX}=1+\frac{BC}{AB}\\ \frac{XF}{AX}=\frac{BC}{AB}\\ \because AB=BC\\ XF=AX\mathrm{...}(1)\end{array}$

No in triangle AFD and XFE:

  $\begin{array}{l}In\Delta \text{AFD and }\Delta \text{XFE}\\ \angle \text{FEX}=\angle FDA(XE||AD)\\ \angle FXE=\angle FAD(XE||AD)\\ \angle F=\angle F(common)\\ by\text{ AAA rule}\\ \Delta \text{AFD~}\Delta \text{XFE}\\ \text{so,}\\ \frac{FA}{FX}=\frac{FD}{FE}=\frac{XE}{AD}\end{array}$

  $\begin{array}{l}\frac{FX+AX}{FX}=\frac{FE+ED}{FE}\\ 1+\frac{AX}{FX}=1+\frac{ED}{FE}\\ \frac{AX}{FX}=\frac{ED}{FE}\\ Byeq(1)\\ \frac{\text{DE}}{\text{EF}}=\text{1}\\ \text{DE}=\text{EF }\end{array}$