Chapter 9.5, Problem 14WE

To determine

To prove $$ $m\angle ATP=\frac{1}{2}m\stackrel{⌢}{ANT}$ When the Centre O lies inside the angleATP

Explanation of Solution

Given Information:

The Centre O lies inside the angleATP as shown below:

Formulas Used:

• Theorem(1)

The measure of an inscribed angle is equal to half the measure of its intercepted arc

• Theorem(2)

The measure of an angle formed by a chord and a tangent is equal to half the measure of the intercepted arc (The Theorem is proved when the Centre Olies on the chord)

• Arc addition postulate

The measure of the arc formed by two adjacent arcs is the sum of the measure of the two arcs

From the diagram given above:

Apply Theorem (2) to the chord ZT since the CentreO lies on arc ZT

.....(1)

Apply Theorem (1) to the arc AZ

  $\angle ATZ=\frac{1}{2}\stackrel{⌢}{AZ}$...... (2)

Apply angle addition postulate to the angle ATP

  $\angle ATP=\angle ATZ+\angle ZTP$

∴ $\angle ZTP=\angle ATP-\angle ATZ$......... (3)

Apply arc addition postulate to the arc ANT

  $\stackrel{⌢}{ANT}=\stackrel{⌢}{AZ}+\stackrel{⌢}{ZNT}$..... (4)

Substitute (1) to the Left Hand Side of (3) and (2) to Right Hand Side of (3)

  $\frac{1}{2}\stackrel{⌢}{ZNT}=\angle ATP-\frac{1}{2}\stackrel{⌢}{AZ}$

  $\Rightarrow \frac{1}{2}\left(\stackrel{⌢}{ZNT}+\stackrel{⌢}{AZ}\right)=\angle ATP$.....(5)

Applying (4) to the (5)

  $\frac{1}{2}\stackrel{⌢}{ANT}=\angle ATP$

Hence the theorem.