5.5 Other Triangle Similarities
We have accepted that if two angles of one triangle are congruent to two angles of a second triangle then the two triangles are similar. There are two other ways to prove that triangles are similar and we can now prove these two as theorems.
- Theorem 5.5:
SAS Similarity Theorem
If an angle of one triangle is congruent to an angle of another triangle and the sides including those angles are proportional, then the triangles are similar.
Given: $\angle
A\cong\angle D$, $\dfrac{AB}{DE}=\dfrac{AC}{DF}$ Prove: $\triangle ABC\sim \triangle DEF$ |
Proof: Choose $X$ on $\overline{AB}$ such that $AX=DE$. Through $X$ draw a line parallel to $\overline{BC}$, intersecting $\overline{AC}$ at $Y$. By Theorem 5.1, we know $\triangle AXY\sim\triangle ABC$, so $\frac{AX}{AB}=\frac{AY}{AC}$. Using our properties of proportions, we then know that $\frac{AX}{AY}=\frac{AB}{AC}$. We were given that $\frac{AB}{DE}=\frac{AC}{DF}$, which can be rewritten as $\frac{AB}{AC}=\frac{DE}{DF}$. Therefore $\frac{AX}{AY}=\frac{DE}{DF}$. Since $AX=DE$, we can conclude that $DF=AY$. Then $\triangle DEF\cong\triangle AXY$ by SAS, and so $\angle E\cong\angle AXY$. Since we introduced $\overline{AX}\parallel\overline{BC}$, we have $\angle AXY\cong\angle B$, and therefore $\angle E\cong \angle B$. Now, with $\angle A\cong\angle D$, we have $\triangle ABC\sim\triangle DEF$ by the AA Similarity Postulate.
Our second theorem is the SSS Similarity Theorem.
- Theorem 5.6:
SSS Similarity Theorem
If three sides of one triangle are proportional to the three corresponding sides of a second triangle, then the two triangles are similar.
Given:
$\dfrac{DE}{AB}=\dfrac{EF}{BC}=\dfrac{DF}{AC}$ Prove: $\triangle DEF\sim\triangle ABC$ |
Proof: Choose $X$ on $\overline{AB}$ such that $AX=DE$. Through $X$ draw a line parallel to $\overline{BC}$, intersecting at $Y$. This makes $\triangle AXY\sim\triangle ABC$ by AA, so $\frac{AX}{AB}=\frac{AY}{AC}=\frac{XY}{BC}.$ Using this proportion, the given proportion, and the fact that $AX=DE$, we can show that $EF=XY$ and $DF=AY$. Then $\triangle DEF\cong\triangle AXY$ by SSS, so $\angle D\cong\angle A$. By using the SAS Similarity Theorem, we can then conclude that $\triangle DEF\sim \triangle ABC$.
Example: Given
$\overline{LP}\perp\overline{EA}$ $N$ is the midpoint of $\overline{LP}$ $P$ and $R$ trisect at $\overline{EA}$ Prove: $\triangle PEN\sim \triangle PAL$ |
Solution: Since $\overline{LP}\perp\overline{EA}$, $\angle NPE$ and $\angle LPA$ are congruent right angles. If $N$ is the midpoint fo $\overline{LP}$, $\frac{NP}{LP}=\frac{1}{2}$. But $P$ and $R$ trisect $\overline{EA}$, so $\frac{EP}{PA}=\frac{1}{2}$. Therefore, $\triangle PEN\sim\triangle PAL$ by the SAS Similarity Theorem.
Example: The sides of one triangle have lengths of 8, 14, and 12 and the sides of another triangle have lengths of 18, 21, and 12. Show that the two triangles are similar.
Solution: Let us determine the ratios of the corresponding sides.
Shortest sides: $\dfrac{8}{12}=\dfrac{2}{3}$ Longest sides: $\dfrac{14}{21}=\dfrac{2}{3}$ Remaining sides: $\dfrac{12}{18}=\dfrac{2}{3}$
Since the ratio is the same for all of the corresponding sides, the two triangles are similar because of the SSS Similarity Theorem.