Chapter 10, Problem 21RP

To determine

To prove : AD and BC are equidistant from F.

Explanation of Solution

Given information :

In circle with centre F, $\overline{FG}\perp \overline{AB}$ and $\overline{EC}\cong \overline{ED}$ .

  

Formula Used:

SAS congruency and AAS congruency.

Calculation :

In $\Delta$ AGE and $\Delta$ BGE

  $\angle AGE=\angle BGE=90°$

EG is common

Since $EG\perp AB$ and AB is a chord, therefore EG bisects AB.

  $\therefore AG=BG$

  $\therefore \Delta AGE\cong \Delta BGE$ [SAS congruence]

  $\therefore \angle EAG=\angle EBG\mathrm{\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}....}(1)$

In $\Delta AGF$ and $\Delta BGF$

AG = BG

  $\angle AGF=\angle BGF=90°$

  $\therefore \Delta AGF\cong \Delta BGF$ [SAS Congruence]

  $\therefore \angle FAG=\angle FBG$    ....(2)

AF = BF     .....(3)

In $\Delta FAP$ and $\Delta FBQ$

AF = BF [Using (3)]

  $\begin{array}{l}\angle FPA=\angle FBQ=90°\\ \angle FPA=\angle EAG-\angle FAG\end{array}$

  $\Rightarrow \angle EBG-\angle FBG=\angle FBQ$ [Using (1) and (2)]

  $\Rightarrow \Delta FAP\cong \Delta FBD$ [AAS congruency]

  $\therefore$ FP = FQ [CPCT]

  $\therefore$ Distance of Ad from F = distance of BC from F.

  $\therefore$ AD and BC are equidistant from F.