Chapter 10.4, Problem 7PSA

a)

To determine

To Prove : $\angle A\cong \angle B$

Answer to Problem 7PSA

xxxx

Explanation of Solution

Given:

  $\overline{CE}\text{ is a common internal tangent to circles A and B at C and E}\text{.}$

Calculation:

  $\begin{array}{l}\text{We know that line through centre is perpendicular}\\ \text{to tangent throughpoint of contact}\text{.}\\ \Rightarrow \angle ACD=98,\angle BED=90°\\ In\Delta ACDand\Delta BED\\ \angle ADC\cong \angle BDE\text{(Vertically opposite angle)}\\ \angle ACD=\angle BED(90°)\\ \text{By}\text{AA - similarity}\\ \angle ACD~\Delta BED\\ \Rightarrow \angle A\cong \angle B(\text{by CPCT})\end{array}$

b)

To determine

To Prove : $\frac{AD}{BD}=\frac{CD}{DE}$

Answer to Problem 7PSA

xxxx

Explanation of Solution

Given:

  $\begin{array}{l}As\Delta ACE~\Delta BED\\ \Rightarrow \frac{AD}{BD}=\frac{CD}{DE}\text{[Ratio of corresponding sides]}\end{array}$