Question

use the figure shown to the right to prove the following.
given: right \\( \triangle a b c \\) with altitude to the hypotenuse \\( \overline{c d} \\)
prove: \\( \frac{a b}{a c}=\frac{a c}{a b}=\frac{b c}{b c}, \frac{a c}{a d}, \frac{b c}{d b}=\frac{}{d b} \\)

1. right \\( \triangle a b c \\) with altitude to the hypotenuse \\( c d \\).
2. \\( \triangle a b c-\triangle a c d, \triangle a b c-\triangle c b d \\)
3. \\( \frac{a b}{a c}=\frac{a c}{a b}=\frac{b c}{b c} \\)
4. \\( \frac{a c}{a d}, \frac{b c}{d b}=\frac{}{d b} \\)
5. given
6. altitude of a right triangle theorem

Explanation:

Step1: Prove similarity of triangles

• For \(\triangle ABC\) and \(\triangle ACD\):
• \(\angle A=\angle A\) (common angle)
• \(\angle ACB = \angle ADC=90^{\circ}\) (given \(CD\) is altitude)
• By AA (Angle - Angle) similarity criterion, \(\triangle ABC\sim\triangle ACD\)
• Then, \(\frac{AB}{AC}=\frac{AC}{AD}=\frac{BC}{CD}\) (corresponding sides of similar triangles are proportional)
• For \(\triangle ABC\) and \(\triangle CBD\):
• \(\angle B=\angle B\) (common angle)
• \(\angle ACB=\angle CDB = 90^{\circ}\) (given \(CD\) is altitude)
• By AA similarity criterion, \(\triangle ABC\sim\triangle CBD\)
• Then, \(\frac{AB}{BC}=\frac{BC}{DB}=\frac{AC}{CD}\) (corresponding sides of similar triangles are proportional)

Step2: Use the similarity results

• From \(\triangle ABC\sim\triangle ACD\), we have \(\frac{AC}{AD}=\frac{AB}{AC}\)
• From \(\triangle ABC\sim\triangle CBD\), we have \(\frac{BC}{DB}=\frac{AB}{BC}\)

Answer:

1. Reason for \(\triangle ABC\sim\triangle ACD\) and \(\triangle ABC\sim\triangle CBD\): AA (Angle - Angle) similarity criterion.
2. Proportions:

• \(\frac{AC}{AD}=\frac{AB}{AC}\) (from \(\triangle ABC\sim\triangle ACD\))
• \(\frac{BC}{DB}=\frac{AB}{BC}\) (from \(\triangle ABC\sim\triangle CBD\))