Chapter 7.3, Problem 25PPS

To determine

To write: a two-column proof of $\Delta ABC\sim \Delta DEF$ .

Answer to Problem 25PPS

The two-column theorem $\Delta ABC\sim \Delta DEF$ is proved.

Explanation of Solution

Given:

  $\Delta ABC\text{ and }\Delta DEF$ are triangles such that $\frac{AB}{DE}=\frac{BC}{EF}\text{ and }\angle B=\angle E$ .

  

Calculation:

According to the question,

  $\Delta ABC\text{ and }\Delta DEF$ are triangles.

Draw $QP\parallel PC$ , so that $QP=EF$ .

Now,

  $\begin{array}{l}\angle AQP=\angle B\text{{corresponding angles}}\\ \angle APQ=\angle C\text{{corresponding angles}}\\ \Delta AQP\sim \Delta ABC\{\text{by }AA-similarity\}\\ \therefore \frac{AQ}{AB}=\frac{QP}{BC}\{\text{by C}\text{.P}\text{.C}\text{.T}\text{.}}\end{array}$ .

  $\begin{array}{l}QP=EF\{\text{given}}\\ \therefore AQ\cdot BC=AB\cdot EF\{\text{cross-multiplication}}\\ \Rightarrow DE\cdot BC=AB\cdot EF\{\because AQ=DE\}\\ AQ=DE\\ \therefore \Delta APQ\cong \Delta DEF\left\{\text{by SAS congruence rule}\right\}\\ \angle APQ=\angle F\{\text{by CPCT}}\\ \Delta ABC\sim \Delta DEF\{\text{by AA-similarity}}\end{array}$

Thus, $\Delta ABC\sim \Delta DEF$

Hence, the two-column theorem is proved.