Geometry: Common Core (15th Edition)

$\triangle NKM$ ~ $\triangle JKL$ using the SAS Similarity Theorem.
We are given the measures of two sides and an included angle in the two triangles. Let's see if the Side-Angle-Side Similarity Theorem can be applied here. The SAS Similarity Theorem states that if two sides in one triangle are proportional to the two sides of another triangle, and the included angle in one triangle is congruent to the included angle of the other triangle, then the two triangles are similar. Let's set up the ratios of corresponding sides in $\triangle NKM$ and $\triangle JKL$: $\frac{JK}{NK} = \frac{15 + 5}{15}$ Evaluate what is in brackets first: $\frac{JK}{NK} = \frac{20}{15}$ Divide both the numerator and denominator by their greatest common factor, $5$: $\frac{JK}{NK} = \frac{4}{3}$ Let's look at $KL$ and $KM$: $\frac{KL}{KM} = \frac{28}{21}$ Divide both the numerator and denominator by their greatest common factor, $7$: $\frac{KL}{KM} = \frac{4}{3}$ Let's look at the corresponding angles in the two triangles, which both measure $90^{\circ}$ because $\overline{JK}$ is perpendicular to $\overline{ML}$: $m \angle JKL ≅ m \angle JKM = 90^{\circ}$ The triangles are similar because two sides in one triangle are proportional to two sides in the other triangle, and the included angles in both triangles are congruent, so $\triangle NKM$ ~ $\triangle JKL$ using the SAS Similarity Theorem.