Chapter 9.3, Problem 15PSB

(a)

To determine

To prove: That the ratios of sides $AD\text{ , }CD\text{ and }CD,BD$ of any two similar triangles are equal i.e. $\frac{AD}{CD}=\frac{CD}{BD}$

Explanation of Solution

Given information :

A circle is given, whose center is $O$ ,

  $CD\perp AB\text{ and }\angle ACB\text{ is a right angle}$

Formula used :

Corresponding sides of similar triangles are equivalent

Proof :

In $\Delta ADC\text{ and }\Delta BDC$ ,

  $\begin{array}{l}\angle ADC\text{=90}°\text{=}\angle BDC\\ DC\text{ is common}\\ \angle DBC\text{=}\angle DAC\end{array}$

Thus, $\Delta ADC\text{ and }\Delta BDC$ are similar.

Therefore,

  $\frac{AD}{CD}=\frac{CD}{BD}$

Hence proved

(b)

To determine

To Show: The ratios of sides $AD\text{ , }ED\text{ and }ED,BD$ are equal i.e. $\frac{AD}{CD}=\frac{CD}{BD}$

Explanation of Solution

Given information :

We have a circle with center $O.$

  $CD\perp AB\text{ and }\angle ACB\text{ is a right angle}\text{.}$

Formula used :

Corresponding sides of similar triangles are equivalent

Proof :

From part (a), we have

  $\frac{AD}{CD}=\frac{CD}{BD}$

We know that the perpendicular is drawn from the center of a circle, bisects the chord $EC$ .

Thus, $CD=ED$

Therefore

  $\frac{AD}{ED}=\frac{ED}{BD}$

Hence proved