Chapter 7.3, Problem 26PPS

To determine

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

Answer to Problem 26PPS

The two-column theorem $\Delta ABC\sim \Delta XYZ$ 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,

  $\begin{array}{l}\Delta ABC\sim \Delta DEF\text{{given}\}\\ \therefore \angle A=\angle D\{\text{by C}\text{.P}\text{.S}\text{.T}\text{.}}\\ \angle B=\angle E\{\text{by C}\text{.P}\text{.S}\text{.T}\text{.}}\\ \text{similarly,}\\ \angle D=\angle A,\angle E=\angle B\\ \Delta DEF\sim \Delta ABC\end{array}$ .

Now,

  $\begin{array}{l}\Delta ABC\sim \Delta DEF,\text{ and }\Delta DEF\sim \Delta ABC\{\text{given}}\\ \therefore \angle A=\angle D\text{ and }\angle B=\angle E\{\text{by CPST} }\\ \angle A=\angle X\text{ and }\angle B=\angle Y\{\text{transitive property of congruent triangles}}\\ \therefore \Delta ABC\sim \Delta XYZ\left\{\text{by AA similarity}\right\}\end{array}$

Thus, $\Delta ABC\sim \Delta XYZ$

Hence, the two-column theorem is proved.