Chapter 7.5, Problem 9CE

a.

To determine

To prove: $\Delta PBE~\Delta PAD$ .

Explanation of Solution

Given information:

ABCD is a parallelogram.

Also,

  $PA=AD=BC=3$ and $AB=DC=CE=7$ .

Proof: 

  $\begin{array}{l}\text{In }\Delta PEB\text{ and }\Delta PAD,\\ \angle PAD=\angle EBP\text{ (AD || BC)}\\ \angle P=\angle P\text{ (Common)}\text{.}\\ \text{Thus, by AA rule,}\\ \Delta PEB~\Delta PAD.\square \end{array}$

b.

To determine

To calculate: Ratio of PB to PA.

Answer to Problem 9CE

  $PB:PA=10:3$.

Explanation of Solution

Given information:

  $\begin{array}{l}PA=AD=BC=3,\text{ and}\\ AB=DC=CE=7.\end{array}$

Calculation: 

Since,

  $\begin{array}{l}\frac{PB}{PA}\Rightarrow \frac{PA+AB}{PA},\\ \Rightarrow \frac{3+7}{3},\\ \Rightarrow \frac{10}{3}.\end{array}$

c.

To determine

To calculate: Ratio of PE to PD.

Answer to Problem 9CE

  

Explanation of Solution

Given information:

  

Also,

  $\begin{array}{l}PA=AD=BC=3,\text{ and}\\ AB=DC=CE=7.\end{array}$

Calculation: 

  $\begin{array}{l}\text{Since, }\Delta PEB~\Delta PAD,\\ \text{Then},\\ \frac{PE}{PD}=\frac{PB}{PA},\\ \Rightarrow \frac{PE}{PD}=\frac{10}{3}.\end{array}$

d.

To determine

To calculate: ratio of $DD\text{'}:EE\text{'}$ .

Answer to Problem 9CE

  $\frac{DD\text{'}}{EE\text{'}}=\frac{3}{10}$ .

Explanation of Solution

Given information: 

  

Also,

  $\begin{array}{l}PA=AD=BC=3,\text{ and}\\ AB=DC=CE=7.\end{array}$

Calculation: 

  $\begin{array}{l}\text{In }\Delta \text{PD'D and }\Delta \text{PE'E}\\ \angle PD\text{'}D=\angle PE\text{'}E(\text{Since }D\text{'}D||E\text{'}E.),\text{ and}\\ \angle PDD\text{'}=\angle PEE\text{'}(\text{Since }DD\text{'}||EE\text{'}.).\\ \text{Thus, by AA rule,}\\ \Delta \text{PD'D ~ }\Delta \text{PE'E,}\\ \Rightarrow \frac{PD\text{'}}{PE\text{'}}=\frac{PD}{PE}=\frac{DD\text{'}}{EE\text{'}}.\\ \text{But, since }\frac{PD}{PE}=\frac{3}{10},\\ \frac{DD\text{'}}{EE\text{'}}=\frac{3}{10}.\end{array}$