below is a picture of a right triangle abc with right angle c along with the point d so that $\overline{cd}$ is perpendicular to $\overline{ab}$.

(image of right triangle abc with right angle at c, cd perpendicular to ab, ad = 5, cd = 12)

a. name all of the triangles that are similar to triangle adc.
$\triangle adc \sim \triangle$ type your answer...
$\sim \triangle$ type your answer...

b. find the lengths of segments $\overline{ac}$, $\overline{cb}$, and $\overline{db}$.
$\overline{ac}$ = type your answer...
$\overline{cb}$ = type your answer...
$\overline{db}$ = type your answer...

Question

below is a picture of a right triangle abc with right angle c along with the point d so that $\overline{cd}$ is perpendicular to $\overline{ab}$.

(image of right triangle abc with right angle at c, cd perpendicular to ab, ad = 5, cd = 12)

a. name all of the triangles that are similar to triangle adc.
$\triangle adc \sim \triangle$ type your answer...
$\sim \triangle$ type your answer...

b. find the lengths of segments $\overline{ac}$, $\overline{cb}$, and $\overline{db}$.
$\overline{ac}$ = type your answer...
$\overline{cb}$ = type your answer...
$\overline{db}$ = type your answer...

Accepted Answer

### Part a
# Explanation:
## Step1: Recall Similar Triangles in Right Triangles
In a right triangle, when an altitude is drawn to the hypotenuse, the two smaller triangles are similar to the original triangle and to each other. For right triangle $ABC$ with right angle $C$ and altitude $CD$ to hypotenuse $AB$, we have:
- $\triangle ADC$ is a right triangle (right angle at $D$).
- $\triangle ABC$ is the original right triangle (right angle at $C$).
- $\triangle CDB$ is the other smaller right triangle (right angle at $D$).

To prove similarity, we use AA (Angle - Angle) criterion.
- For $\triangle ADC$ and $\triangle ABC$: $\angle ADC=\angle ACB = 90^{\circ}$ and $\angle A$ is common to both triangles. So, by AA, $\triangle ADC\sim\triangle ABC$.
- For $\triangle ADC$ and $\triangle CDB$: $\angle ADC=\angle CDB = 90^{\circ}$, and $\angle ACD=\angle B$ (because $\angle A+\angle ACD = 90^{\circ}$ and $\angle A+\angle B=90^{\circ}$, so $\angle ACD=\angle B$). So, by AA, $\triangle ADC\sim\triangle CDB$.

# Answer:
$\triangle ADC\sim\triangle ABC$ $\sim\triangle CDB$

### Part b
# Explanation:
## Step1: Find $AC$
In right triangle $ADC$, we know $AD = 5$ and $CD=12$. Using the Pythagorean theorem $AC^{2}=AD^{2}+CD^{2}$ (since $\triangle ADC$ is a right triangle with legs $AD$ and $CD$ and hypotenuse $AC$).
$$AC=\sqrt{AD^{2}+CD^{2}}=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13$$

## Step2: Find $AC$ (Alternative - Using Similar Triangles for $CB$ and $DB$)
Since $\triangle ADC\sim\triangle CDB$, the ratios of corresponding sides are equal. That is, $\frac{AD}{CD}=\frac{CD}{DB}$ (corresponding sides of similar triangles). We know $AD = 5$ and $CD = 12$. Let $DB=x$. Then $\frac{5}{12}=\frac{12}{x}$, cross - multiplying gives $5x=144$, so $x=\frac{144}{5}=28.8$. But we can also use the similarity between $\triangle ADC$ and $\triangle ABC$ to find $AB$ first.

Since $\triangle ADC\sim\triangle ABC$, $\frac{AD}{AC}=\frac{AC}{AB}$. We know $AD = 5$ and $AC = 13$. Let $AB=y$. Then $\frac{5}{13}=\frac{13}{y}$, cross - multiplying gives $5y = 169$, so $y=\frac{169}{5}=33.8$. But $AB=AD + DB$, so $DB=AB - AD=33.8 - 5 = 28.8=\frac{144}{5}$.

Now, to find $CB$, since $\triangle ADC\sim\triangle CDB$, $\frac{CD}{CB}=\frac{AD}{CD}$. We know $CD = 12$ and $AD = 5$. Let $CB = z$. Then $\frac{12}{z}=\frac{5}{12}$, cross - multiplying gives $5z=144$, so $z=\frac{144}{5}=28.8$? Wait, no. Wait, from $\triangle ABC$, using Pythagorean theorem: $AB=\frac{AC^{2}}{AD}=\frac{13^{2}}{5}=\frac{169}{5}$, $CB=\sqrt{AB^{2}-AC^{2}}=\sqrt{(\frac{169}{5})^{2}-13^{2}}=\sqrt{\frac{28561}{25}-\frac{4225}{25}}=\sqrt{\frac{28561 - 4225}{25}}=\sqrt{\frac{24336}{25}}=\frac{156}{5}=31.2$? Wait, I made a mistake earlier. Let's correct it.

Wait, in right triangle $ADC$, $AD = 5$, $CD = 12$, so $AC=\sqrt{5^{2}+12^{2}} = 13$ (correct). Since $\triangle ADC\sim\triangle CDB$, $\angle A=\angle BCD$ and $\angle ACD=\angle B$. Also, $\triangle ABC\sim\triangle ADC$, so $\frac{AC}{AB}=\frac{AD}{AC}$, so $AB=\frac{AC^{2}}{AD}=\frac{13^{2}}{5}=\frac{169}{5} = 33.8$. Then $DB=AB - AD=\frac{169}{5}-5=\frac{169 - 25}{5}=\frac{144}{5}=28.8$. And $CB$: in $\triangle ABC$, using Pythagorean theorem $CB=\sqrt{AB^{2}-AC^{2}}=\sqrt{(\frac{169}{5})^{2}-13^{2}}=\sqrt{\frac{28561}{25}-\frac{4225}{25}}=\sqrt{\frac{24336}{25}}=\frac{156}{5}=31.2$. Wait, another way: since $\triangle ADC\sim\triangle CDB$, $\frac{CD}{CB}=\frac{AD}{CD}$, so $CB=\frac{CD^{2}}{AD}=\frac{12^{2}}{5}=\frac{144}{5}=28.8$? No, that's wrong. Wait, the correct ratio for similar triangles $\triangle ADC$ and $\triangle CDB$ is $\frac{AD}{CD}=\frac{CD}{DB}=\frac{AC}{CB}$. So $\frac{AD}{CD}=\frac{5}{12}$, $\frac{CD}{DB}=\frac{12}{DB}$, so $\frac{5}{12}=\frac{12}{DB}$ gives $DB=\frac{144}{5}=28.8$. And $\frac{AC}{CB}=\frac{5}{12}$, since $AC = 13$, then $CB=\frac{12\times13}{5}=\frac{156}{5}=31.2$. Wait, let's use Pythagorean theorem on $\triangle ABC$. $AB=AD + DB=5+\frac{144}{5}=\frac{25 + 144}{5}=\frac{169}{5}$. Then $BC=\sqrt{AB^{2}-AC^{2}}=\sqrt{(\frac{169}{5})^{2}-13^{2}}=\sqrt{\frac{28561 - 4225}{25}}=\sqrt{\frac{24336}{25}}=\frac{156}{5}=31.2$. And $AC = 13$ (from Pythagorean theorem in $\triangle ADC$: $5 - 12-13$ triangle).

## Step1 (Correct for $AC$):
In right triangle $ADC$, by Pythagorean theorem:
$$AC=\sqrt{AD^{2}+CD^{2}}=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13$$

## Step2 (For $CB$):
Since $\triangle ADC\sim\triangle CDB$, $\frac{CD}{CB}=\frac{AD}{CD}$ (corresponding sides). So $CB=\frac{CD^{2}}{AD}=\frac{12^{2}}{5}=\frac{144}{5}=28.8$? No, wait, the correct correspondence is $\triangle ADC\sim\triangle CDB$ means $\angle A=\angle BCD$, $\angle ADC=\angle CDB = 90^{\circ}$, so the ratio of sides: $\frac{AD}{CD}=\frac{CD}{DB}=\frac{AC}{CB}$. So $\frac{AC}{CB}=\frac{AD}{CD}=\frac{5}{12}$, and since $AC = 13$, then $CB=\frac{12\times13}{5}=\frac{156}{5}=31.2$.

## Step3 (For $DB$):
Since $AB=AD + DB$, and from $\triangle ADC\sim\triangle ABC$, $\frac{AD}{AC}=\frac{AC}{AB}$, so $AB=\frac{AC^{2}}{AD}=\frac{13^{2}}{5}=\frac{169}{5}$. Then $DB=AB - AD=\frac{169}{5}-5=\frac{169 - 25}{5}=\frac{144}{5}=28.8$

# Answer:
$\overline{AC}=\boldsymbol{13}$

$\overline{CB}=\boldsymbol{\frac{144}{5}=28.8}$ (or $\frac{156}{5}=31.2$ - Wait, no, let's re - check. The correct 3 - 4 - 5 triangle scaled? Wait, 5 - 12 - 13 is a Pythagorean triple. Then $\triangle ADC$ is 5 - 12 - 13. Then $\triangle CDB$ should be 12 - 31.2 - 33.6? No, wait, the correct formula for the altitude to the hypotenuse: $CD^{2}=AD\times DB$, so $12^{2}=5\times DB$, so $DB=\frac{144}{5}=28.8$. Then $AB=AD + DB=5 + 28.8 = 33.8=\frac{169}{5}$. Then $AC = 13$, $CB=\sqrt{AB^{2}-AC^{2}}=\sqrt{(\frac{169}{5})^{2}-13^{2}}=\sqrt{\frac{28561-4225}{25}}=\sqrt{\frac{24336}{25}}=\frac{156}{5}=31.2$. And from $\triangle CDB$, $CB=\sqrt{CD^{2}+DB^{2}}=\sqrt{12^{2}+28.8^{2}}=\sqrt{144 + 829.44}=\sqrt{973.44}=31.2$, which matches. And $AC = 13$ (from $\triangle ADC$: $\sqrt{5^{2}+12^{2}} = 13$).

So:
$\overline{AC}=\boldsymbol{13}$
$\overline{CB}=\boldsymbol{\frac{156}{5}=31.2}$ (or $31.2$)
$\overline{DB}=\boldsymbol{\frac{144}{5}=28.8}$ (or $28.8$)

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QUESTION IMAGE

Question

Explanation:

Part a

Step1: Recall Similar Triangles in Right Triangles

Step1: Find \(AC\)

Step2: Find \(AC\) (Alternative - Using Similar Triangles for \(CB\) and \(DB\))

Step1 (Correct for \(AC\)):

Step2 (For \(CB\)):

Answer:

Part b