Exercise 43 Page 590

The length of $\Seg{KP}$ is $\colIII{5}.$ This means that the length of $\Seg{MP}$ is $\textcolor{darkviolet}{10}.$ Let's add this information to our picture.

Next, as $\triangle MNK$ is a right triangle, we can use the Pythagorean Theorem to evaluate the length of $\Seg{MN}.$ According to this theorem, the sum of the squared legs of a triangle is equal to its squared hypotenuse.

The length of $\Seg{MN}$ is $\textcolor{teal}{12}.$ Using this information, we will evaluate the length of $\Seg{ML}.$ To do this, let's again use the Pythagorean Theorem, as $\triangle MNL$ is also a right triangle.

The length of $\Seg{ML}$ is $\textcolor{#E67700}{20}.$ Finally, we still need to evaluate lengths of $\Seg{MR}$ and $\Seg{PR}.$ Let's look at the picture for the last time.

Since we are given that $\Seg{PR}$ is parallel to $\Seg{KL},$ angles $\angle MRP$ and $\angle MLK$ are congruent as well as $\angle MPR$ and $\angle MKL.$ This means that $\triangle MPR$ and $\triangle MKL$ are similar by Angle-Angle Similarity Theorem.

We can write that corresponding sides are proportional. \begin{gathered} \dfrac{PR}{KL}=\dfrac{MP}{MK}=\dfrac{MR}{ML} \end{gathered} Let's solve the above proportion by substituting appropriate side lengths. We will start with the left and the middle ratio.

The length of $\Seg{PR}$ is $16\frac{2}{3}.$ Finally, we will find the length of $\Seg{MR}$ by evaluating the middle and the right ratio.

The length of $\Seg{MR}$ is $16\frac{2}{3}.$ We can present our answers in a table.