5.3 Similar Triangles
We have been able to prove that two triangles were congruent using a number of postulates. We have a special postulate that we accept to establish the similarity of triangles.
- AA Postulate: If two angles of one triangle are congruent to two angles of a second triangle, then the triangles are similar.
Solution:
\begin{array}{ l l | l}
\text{Proof:} & \text{Statements} & \text{Reasons} \\
\hline
& 1. ABCD\text{ is a parallelogram} & 1. \text{ Given} \\
& 2. \overline{AB}\parallel\overline{CD} & 2. \text{ Def. of parallelogram} \\
A\;& 3. \angle CDF\cong\angle E & 3. \; P\rightarrow AIC\\
A\;& 4. \angle DFC\cong\angle EFB & 4. \text{ Vertical Angle Theorem}\\
& 5. \triangle BFE\sim\triangle CFD & 5. \text{ AA Postulate}\\
\end{array}
The AA Postulate leads us to an interesting result. If a line is parallel to one side of a triangle and intersects the other two sides in two points, it forms another triangle. It appears as if $\triangle ABC\sim\triangle APR$. Drag the vertices of $\triangle ABC$ or point $P$ in the figure below to see if this remains true.
Let’s see if we can prove this as a theorem.
- Theorem 5.1:
If a line is parallel to one side of a triangle and intersects the
other two sides in two points, then the line cuts off a triangle that
is similar to the given triangle.
Prove: $\triangle APR\sim\triangle ABC$
Proof: Since parallel lines $\overline{PR}$ and $\overline{BC}$ are cut by transversal $\overline{APB}$, corresponding angles $\angle APR$ and $\angle ABC$ are congruent. Since $\angle A\cong \angle A$, then $\triangle APR\sim\triangle ABC$ by the AA Postulate.
Example: Given $\triangle ABC$ with $\overline{PR}\parallel \overline{BC}$; $AR=14$, $PR=12$, $BC=15$, and $PB=4$. Find $AP$, $AB$, and $AC$. |
We use Theorem 5.1 in a slightly different way in the following example.
Example: In trapezoid $ABCD$, $\overline{EF}\parallel\overline{CD}$; $AB=24$ and $CD=54$. If $AE:ED=3:7$, find the length of $\overline{EF}$ | |
Solution: Draw diagonal
$\overline{DB}$. We can then use Theorem 5.1 in each of the
triangles
$\triangle ABD$ and $\triangle BDC$. In $\triangle ABD$, we have
the
proportion $$\frac{EG}{AB}=\frac{7}{10}.$$ This gives
$EG=\frac{7}{10}\cdot24=\frac{168}{10}$. In $\triangle BDC$, we
have the proportion $$\frac{GF}{CD}=\frac{3}{10}.$$ This least to
$GF=\frac{3}{10}\cdot 54=\frac{162}{10}$Adding the results, we have
$$EF=\frac{168}{10}+\frac{162}{10}=33.$$ |