Chapter 4.4, Problem 17PSB

To determine

To Verify: $AF\perp BC.$

Explanation of Solution

Given information: $DE=EC,DB\perp DF,EC\perp EF,BF=FC.$

Formula used: congruency of triangles, linear pair axiom.

Proof:

  $DE=EC,DB\perp DF,EC\perp EF,BF=FC(\text{given})$

  $\begin{array}{l}\therefore \angle BDF=\angle CEF(\text{Each}\text{equal}\text{to}{\text{90}}^{\text{o}})\\ \therefore \Delta BDF\cong \Delta CEF(\text{RHS})\end{array}$

  $\begin{array}{l}\therefore \angle BDF=\angle CEF\text{(CPCT)}\mathrm{.....}\text{(1)}\\ \angle B=\angle C\text{(CPCT)}\end{array}$

  $\begin{array}{l}\therefore AB=AC\text{(Sides opposite to equal angles are equal)}\\ \therefore AB-DB=AC-CE\text{(}\because \text{DB=CE)}\end{array}$

  $\begin{array}{l}\therefore \text{AD=AE}\\ AF=AF(\text{Reflective}\text{property)}\end{array}$

  $\begin{array}{l}\therefore \Delta ADF\cong \Delta AEF(\text{RHS})\\ \therefore \angle AFD=\angle AFE(CPCT)\mathrm{....}(2)\end{array}$

  $\begin{array}{l}\text{Adding}\text{equations}\text{(1)}\text{and}\text{(2)}\\ \angle AFD+\angle BFD=\angle AFE+\angle CFE\end{array}$

  $\begin{array}{l}\angle AFB=\angle AFC\\ \angle AFB+\angle AFC={180}^{\circ }(\text{linear}\text{pair})\end{array}$

  $\therefore \angle AFB={90}^{\circ }=\angle AFC.$