Chapter 9.6, Problem 32WE

To determine

To prove: the relationship between $\stackrel{⌢}{XA}$ and $\stackrel{⌢}{XB}$ .

Explanation of Solution

Given information:

  $PK=PT$ ; $PT$ is tangent; and $PBA$ is secant.

Proof: 

  

  $\begin{array}{l}PT=PX,\\ \angle PTK=\angle PKT,\\ \angle PKT=\angle AKX(\text{Vertically opposite angle})\mathrm{...}(1).\end{array}$

Now, draw a line PX.

  $\begin{array}{l}\text{In }\Delta PTK\text{ and }\Delta PKX\\ PT=PX\text{ (tangent to the circle),}\\ \angle PXK=\angle PTK(\text{same point of toching the circle from same distance}),\text{ and}\\ PK=PK(\text{common}).\\ \text{So, by SAS rule,}\\ \Delta PTK\cong \Delta PKX.\\ \text{Thus, by congurent property, }\\ \angle PKT=\angle PKX\mathrm{...}(2)\end{array}$

Now, by eq - (1) and eq - (2), we have:

  $\stackrel{⌢}{AX}=\stackrel{⌢}{BX}$ . $\square$