Chapter 10.6, Problem 21PSB

To determine

Give the explanation for the given proof.

Explanation of Solution

Given:

The given diagram is

  

  $\overline{AC}$ is tangent.

The given proof is $\Delta ADC\sim \Delta BDA$ .

Calculation:

The angle inscribed in a semicircle is a right angle.

  $\begin{array}{l}m\angle ADB={90}^{°}\\ m\angle ADC={180}^{°}-{90}^{°}={90}^{°}\\ m\angle ADB=m\angle ADC={90}^{°}\end{array}$

Tangent chord angle $=\frac{1}{2}$ intercepted arcs

  $m\angle DAC=\frac{1}{2}Arc\left(AD\right)$

Inscribed angle $=\frac{1}{2}$ intercepted arcs

  $m\angle ABD=\frac{1}{2}Arc\left(AD\right)$

Now, $m\angle DAC=m\angle ABD$

Hence from $AA$ similarity $\Delta ADC\sim \Delta BDA$ .