Chapter 11.7, Problem 32WE

To determine

Prove :

Area of $\Delta XYZ=\frac{1}{7}(\text{area of }\Delta \text{ABC)}$

Explanation of Solution

Given:

  $\Delta ABC$ is equilateral ; $\frac{AP}{PB}=\frac{BQ}{QC}=\frac{CR}{RA}=\frac{2}{1}$

  

Calculation:

Consider $\Delta CAP,\Delta ABQ\text{ and }\Delta BCR$ :

  $\angle CAP\cong \angle ABQ\cong \angle BCR\mathrm{........}[\text{equilateral triangle]}$

  $\overline{CA}=\overline{AB}=\overline{BC}\mathrm{..................}[\text{equilateral triangle]}$

  $\overline{AP}=\overline{BQ}=\overline{CR}$

So, by SAS Postulate , $\Delta CAP\cong \Delta ABQ\cong \Delta BCR$

So, $\overline{CP}=\overline{AQ}=\overline{BR}$

  $\begin{array}{l}\angle QAB\cong \angle RBC\cong \angle PCA\\ \Rightarrow \angle YAR\cong \angle ZBP\cong \angle XCQ\end{array}$

  $\begin{array}{l}\angle CPA\cong \angle AQB\cong \angle BRC\\ \Rightarrow \angle ZPB\cong \angle XQC\cong \angle YRA\end{array}$

Also , $\overline{RA}=\overline{QC}=\overline{PB}$

So, by AAS Postulate , $\Delta RAY\cong \Delta QCX\cong \Delta PBZ$

So,

  $\begin{array}{l}\angle RYA\cong \angle QXC\cong \angle PZB\\ \Rightarrow \angle XYZ\cong \angle YXZ\cong \angle XZY\mathrm{..........}[\text{vertically opposite angles]}\end{array}$

Hence , XYZ is equilateral triangle.

Also, $YZ=\frac{1}{\sqrt{7}}AB$

So,

  $\begin{array}{l}\frac{\text{area of }\Delta XYZ}{\text{area of }\Delta \text{ABC}}=\frac{\frac{\sqrt{3}}{4}Y{Z}^2}{\frac{\sqrt{3}}{4}A{B}^2}=\frac{Y{Z}^2}{A{B}^2}=\frac{{\left(\frac{1}{\sqrt{7}}AB\right)}^2}{A{B}^2}=\frac{\frac{1}{7}A{B}^2}{A{B}^2}=\frac{1}{7}\\ \Rightarrow \frac{\text{area of }\Delta XYZ}{\text{area of }\Delta \text{ABC}}=\frac{1}{7}\\ \Rightarrow \text{area of }\Delta XYZ=\frac{1}{7}\left(\text{area of }\Delta \text{ABC}\right)\end{array}$

Hence proved.