Chapter 7.5, Problem 23WE

To determine

To Show: The triangles $\text{Δ}ATB$ and $\text{Δ}{A}^{\prime }T{B}^{\prime }$ are similar.

Explanation of Solution

Given information:

Consider the figure of parallelogram WXYZ.

  

Consider the given figure:

In ?WTB and ?YTB',

  $\angle WTB=\angle YTB\text{'}$ (Vertically opposite angles )

  $\angle WBT=\angle YB\text{'}T$ (Alternate interior angles)

This implies that $\Delta WTB~\Delta YTB\text{'}$

Since the corresponding sides of similar triangles are proportional.

  $\frac{BT}{B^{\prime }T}=\frac{TW}{TY}$

In a similar way, first prove that ?WTA and ?YTA' are similar.

Then it follows that,

  $\frac{AT}{A^{\prime }T}=\frac{TW}{TY}$

  $\frac{BT}{B^{\prime }T}=\frac{AT}{A^{\prime }T}=\frac{TW}{TY}$

Also $\angle ATB=\angle A^{\prime }T{B}^{\prime }$

Therefore the triangles $\text{Δ}ATB$ and $\text{Δ}{A}^{\prime }T{B}^{\prime }$ are similar using SAS similarity theorem.