Chapter 2, Problem 5RP

To determine

To Conclude: In the given figure $\overline{PT}\cong \overline{PX}$ means they both are congruent

Explanation of Solution

Given information : In the figure we are given $\overline{ST}\cong \overline{WX}$ and $\overline{RV}\text{ , }\overline{SW}$ trisect $\overline{PT}\text{ , }\overline{PX}$

Formula used : If two segments are congruent, then their like multiples are also congruent.

Proof: As per the question we have $\overline{ST}\cong \overline{WX}$ .As both are congruent it means their multiples also will be congruent. So, we can write it as:

  $3\overline{ST}\cong \text{ 3 }\overline{WX}\mathrm{.........................................................................}\text{(i)}$

Now $\overline{RV}\text{ , }\overline{SW}$ trisect $\overline{PT}\text{ , }\overline{PX}$ which means they divide both of them in three equal parts:

  $\begin{array}{l}\overline{PV}\text{ = }\overline{VW}\text{ = }\overline{WX}\mathrm{.................................................}\text{(ii)}\\ \\ \overline{PR}\text{ = }\overline{RS}\text{ = }\overline{ST}\mathrm{.................................................}\text{(iii)}\end{array}$

Consider $\overline{PT}$ as per figure it can be written as:

  $\begin{array}{l}\overline{PT}\text{ = }\overline{PR}\text{ + }\overline{RS}\text{ + }\overline{ST}\\ \\ \Rightarrow \overline{PT}\text{ = }\overline{ST}\text{ + }\overline{ST}\text{ + }\overline{ST}\mathrm{...................................}\text{(from iii)}\\ \\ \Rightarrow \overline{PT}\text{ =3 }\overline{ST}\mathrm{......................................................}\text{(iv)}\end{array}$

Consider $\overline{PX}$ as per figure it can be written as:

  $\begin{array}{l}\overline{PX}\text{ = }\overline{PV}\text{ + }\overline{VW}\text{ + }\overline{WX}\\ \\ \Rightarrow \overline{PX}\text{ = }\overline{WX}\text{ + }\overline{WX}\text{ + }\overline{WX}\mathrm{...................................}\text{(from ii)}\\ \\ \Rightarrow \overline{PX}\text{ =3 }\overline{WX}\mathrm{......................................................}\text{(v)}\end{array}$

Plugging in values from (iv) and (v) in(i) we get:

  $\begin{array}{l}\text{3 }\overline{ST}\cong \text{ 3 }\overline{WX}\\ \\ \Rightarrow \overline{PT}\cong \overline{PX}\end{array}$

Which proves the required result.