Chapter 7.5, Problem 21WE

To determine

To Prove: For the given figure , $\Delta RST~\Delta R\text{'}S\text{'}T\text{'}$ .

Explanation of Solution

Given information:

  $\begin{array}{l}O{R}^{\prime }=2\cdot OR\hfill \\ O{S}^{\prime }=2\cdot OS\hfill \\ O{T}^{\prime }=2\cdot OT\hfill \end{array}$

  

Refer to the figure provided in the textbook.

  

Note that,

  $\begin{array}{l}O{R}^{\prime }=2\cdot OR\hfill \\ O{S}^{\prime }=2\cdot OS\hfill \\ O{T}^{\prime }=2\cdot OT\hfill \end{array}$

Here R is the midpoint of OR' , S is the midpoint of OS' , and T is the midpoint of OT' .

  $\frac{R^{\prime }{S}^{\prime }}{RS}=\frac{S{T}^{\prime }}{ST}=\frac{T^{\prime }{R}^{\prime }}{TR}$

If the corresponding sides of two triangles are in the same ratio, then the triangles are congruent.

Therefore, it follows that $\Delta RST~\Delta R\text{'}S\text{'}T\text{'}$ .