To determine why the triangles are similar, allowing you to write the proportions [tex]\(\frac{c}{a} = \frac{a}{f}\)[/tex] and [tex]\(\frac{c}{b} = \frac{b}{e}\)[/tex], let's examine the triangles involved.
Consider a right triangle [tex]\( \triangle ABC \)[/tex] with [tex]\( \angle C = 90^\circ \)[/tex]. If you draw the altitude from [tex]\( C \)[/tex] perpendicular to [tex]\( AB \)[/tex], this altitude will create two smaller right triangles, [tex]\( \triangle ACD \)[/tex] and [tex]\( \triangle CBD \)[/tex], where [tex]\( D \)[/tex] is the point where the altitude meets [tex]\( AB \)[/tex].
The resulting triangles, [tex]\( \triangle ACD \)[/tex] and [tex]\( \triangle CBD \)[/tex], are similar to the original triangle [tex]\( \triangle ABC \)[/tex] because of the Right Triangle Altitude Theorem, which states:
- The altitude drawn to the hypotenuse of a right triangle creates two triangles that are similar to the original triangle and to each other.
Therefore, we can state that the triangles are similar based on the Right Triangle Altitude Theorem.
In other words, the similarity allows us to establish the proportions:
- For [tex]\(\triangle ACD\)[/tex] and [tex]\(\triangle ABC\)[/tex]:
[tex]\[
\frac{AC}{AB} = \frac{AD}{AC} \implies \frac{a}{c} = \frac{f}{a} \implies \frac{c}{a} = \frac{a}{f}
\][/tex]
- For [tex]\(\triangle CBD\)[/tex] and [tex]\(\triangle ABC\)[/tex]:
[tex]\[
\frac{BC}{AB} = \frac{BD}{BC} \implies \frac{b}{c} = \frac{e}{b} \implies \frac{c}{b} = \frac{b}{e}
\][/tex]
Thus, the correct theorem that justifies the similarity of these triangles and the proportions [tex]\(\frac{c}{a} = \frac{a}{f}\)[/tex] and [tex]\(\frac{c}{b} = \frac{b}{e}\)[/tex] is:
The Right Triangle Altitude Theorem
