the figure shows right triangle abc with altitude $ \overline{cd} $ where $ \triangle acd \sim \triangle cbd \sim \triangle abc $.
move the options to the spaces to prove the pythagorean theorem.
| statement | reason |
| --- | --- |
| $ \triangle acd \sim \triangle cbd \sim \triangle abc $ | given |
| $ \frac{a}{c} = \square, \frac{b}{c} = \square $ | corresponding sides of similar triangles are proportional. |
| $ a^2 = cx, b^2 = cy $ | |
| $ a^2 + b^2 = cx + cy $ | |
| $ a^2 + b^2 = c(x + y) $ | distributive property |
| $ a^2 + b^2 = c^2 $ | |
options: $ \frac{x}{a} $, $ \frac{y}{b} $, $ \frac{z}{a} $, $ \frac{z}{b} $, addition property, multiplication property, substitution property, pythagorean theorem

Question

the figure shows right triangle abc with altitude $ \overline{cd} $ where $ \triangle acd \sim \triangle cbd \sim \triangle abc $.
move the options to the spaces to prove the pythagorean theorem.
| statement | reason |
| --- | --- |
| $ \triangle acd \sim \triangle cbd \sim \triangle abc $ | given |
| $ \frac{a}{c} = \square, \frac{b}{c} = \square $ | corresponding sides of similar triangles are proportional. |
| $ a^2 = cx, b^2 = cy $ |  |
| $ a^2 + b^2 = cx + cy $ |  |
| $ a^2 + b^2 = c(x + y) $ | distributive property |
| $ a^2 + b^2 = c^2 $ |  |
options: $ \frac{x}{a} $, $ \frac{y}{b} $, $ \frac{z}{a} $, $ \frac{z}{b} $, addition property, multiplication property, substitution property, pythagorean theorem

Accepted Answer

# Explanation:
## Step1: Analyze the first proportion
For similar triangles $ \triangle ACD \sim \triangle ABC $, the ratio of corresponding sides: $ \frac{a}{c}=\frac{x}{a} $? Wait, no, let's look at the sides. In $ \triangle ACD $ and $ \triangle ABC $, side $ a $ (CB) corresponds to side $ c $ (AB), and side $ x $ (DB) corresponds to side $ a $ (CB)? Wait, maybe better to look at the labels. The right triangle $ ABC $ with altitude $ CD $, so $ AC = b $, $ BC = a $, $ AB = c $, $ AD = y $, $ DB = x $, $ CD = z $. So $ \triangle ACD \sim \triangle CBD \sim \triangle ABC $. So for $ \triangle ACD $ and $ \triangle ABC $: $ \frac{AC}{AB}=\frac{AD}{AC} $, so $ \frac{b}{c}=\frac{y}{b} $? Wait, no, the first ratio is $ \frac{a}{c}=\frac{x}{a} $ (since $ \triangle CBD \sim \triangle ABC $: $ \frac{CB}{AB}=\frac{DB}{CB} $, so $ \frac{a}{c}=\frac{x}{a} $), and for $ \triangle ACD \sim \triangle ABC $: $ \frac{AC}{AB}=\frac{AD}{AC} $, so $ \frac{b}{c}=\frac{y}{b} $? Wait, the first blank in $ \frac{a}{c}=\square $, and the second in $ \frac{b}{c}=\square $. Wait, the options are $ \frac{x}{a}, \frac{y}{b}, \frac{z}{a}, \frac{z}{b} $. Wait, maybe $ \triangle CBD \sim \triangle ABC $: corresponding sides $ CB/AB = BD/CB $, so $ a/c = x/a $, so first blank is $ x/a $? No, the first ratio is $ \frac{a}{c}=\square $, so $ \square = \frac{a^2}{c} $? No, the options are $ \frac{x}{a}, \frac{y}{b}, \frac{z}{a}, \frac{z}{b} $. Wait, maybe $ \triangle ACD \sim \triangle CBD $: no, the given similarity is $ \triangle ACD \sim \triangle CBD \sim \triangle ABC $. So $ \triangle ACD \sim \triangle ABC $: $ AC/AB = AD/AC = CD/BC $, so $ b/c = y/b = z/a $. $ \triangle CBD \sim \triangle ABC $: $ CB/AB = BD/CB = CD/AC $, so $ a/c = x/a = z/b $. So the first ratio $ \frac{a}{c} = \frac{x}{a} $ (from $ \triangle CBD \sim \triangle ABC $: $ \frac{CB}{AB} = \frac{BD}{CB} $, so $ \frac{a}{c} = \frac{x}{a} $), so the first blank is $ \frac{x}{a} $? Wait, no, the first expression is $ \frac{a}{c} = \square $, so $ \square = \frac{a^2}{c} $? No, the options are $ \frac{x}{a}, \frac{y}{b}, \frac{z}{a}, \frac{z}{b} $. Wait, maybe I got the sides wrong. Let's re-express:

For $ \triangle CBD \sim \triangle ABC $:

- $ \angle CDB = \angle ACB = 90^\circ $
- $ \angle B $ is common, so similarity. Thus, $ \frac{CB}{AB} = \frac{BD}{CB} $, so $ \frac{a}{c} = \frac{x}{a} $, so $ a^2 = cx $.

For $ \triangle ACD \sim \triangle ABC $:

- $ \angle CDA = \angle ACB = 90^\circ $
- $ \angle A $ is common, so similarity. Thus, $ \frac{AC}{AB} = \frac{AD}{AC} $, so $ \frac{b}{c} = \frac{y}{b} $, so $ b^2 = cy $.

So the first blank (in $ \frac{a}{c} = \square $) is $ \frac{x}{a} $ (since $ \frac{a}{c} = \frac{x}{a} $ from $ \triangle CBD \sim \triangle ABC $), and the second blank (in $ \frac{b}{c} = \square $) is $ \frac{y}{b} $ (since $ \frac{b}{c} = \frac{y}{b} $ from $ \triangle ACD \sim \triangle ABC $).

Then, the next step: $ a^2 = cx $ and $ b^2 = cy $, so adding them: $ a^2 + b^2 = cx + cy $ (addition property), then factoring: $ a^2 + b^2 = c(x + y) $ (distributive property), and since $ x + y = c $ (because $ AD + DB = AB $), so substitution property gives $ a^2 + b^2 = c^2 $.

So first blank: $ \frac{x}{a} $, second blank: $ \frac{y}{b} $. Then the reason for $ a^2 = cx, b^2 = cy $ is multiplication property (cross-multiplying the proportions: $ \frac{a}{c} = \frac{x}{a} $ implies $ a^2 = cx $ by multiplication property, same for $ \frac{b}{c} = \frac{y}{b} $ implies $ b^2 = cy $). Then $ a^2 + b^2 = cx + cy $ is addition property (adding the two equations). Then $ a^2 + b^2 = c(x + y) $ is distributive, then $ x + y = c $ (since $ AD + DB = AB $), so substitution property gives $ a^2 + b^2 = c^2 $.

## Step1: Fill the first blank
The proportion $ \frac{a}{c} = \frac{x}{a} $ (from $ \triangle CBD \sim \triangle ABC $: $ CB/AB = BD/CB $), so first blank is $ \frac{x}{a} $.

## Step2: Fill the second blank
The proportion $ \frac{b}{c} = \frac{y}{b} $ (from $ \triangle ACD \sim \triangle ABC $: $ AC/AB = AD/AC $), so second blank is $ \frac{y}{b} $.

## Step3: Reason for $ a^2 = cx, b^2 = cy $
Cross-multiplying the proportions $ \frac{a}{c} = \frac{x}{a} $ and $ \frac{b}{c} = \frac{y}{b} $ uses the multiplication property (multiplying both sides by $ c $ and $ a $ for the first, $ c $ and $ b $ for the second).

## Step4: Reason for $ a^2 + b^2 = cx + cy $
Adding the two equations $ a^2 = cx $ and $ b^2 = cy $ uses the addition property.

## Step5: Reason for $ a^2 + b^2 = c^2 $
Since $ x + y = c $ (because $ AD + DB = AB $), substituting $ x + y $ with $ c $ uses the substitution property.

So filling the blanks:

First blank (in $ \frac{a}{c} = \square $): $ \frac{x}{a} $

Second blank (in $ \frac{b}{c} = \square $): $ \frac{y}{b} $

Reason for $ a^2 = cx, b^2 = cy $: Multiplication property

Reason for $ a^2 + b^2 = cx + cy $: Addition property

Reason for $ a^2 + b^2 = c^2 $: Substitution property (since $ x + y = c $)

# Answer:
First blank: $ \frac{x}{a} $

Second blank: $ \frac{y}{b} $

Reason for $ a^2 = cx, b^2 = cy $: Multiplication property

Reason for $ a^2 + b^2 = cx + cy $: Addition property

Reason for $ a^2 + b^2 = c^2 $: Substitution property (since $ x + y = c $)

So the filled table:

| Statement | Reason |
| --- | --- |
| $ \triangle ACD \sim \triangle CBD \sim \triangle ABC $ | Given |
| $ \frac{a}{c} = \frac{x}{a}, \frac{b}{c} = \frac{y}{b} $ | Corresponding sides of similar triangles are proportional. |
| $ a^2 = cx, b^2 = cy $ | Multiplication property |
| $ a^2 + b^2 = cx + cy $ | Addition property |
| $ a^2 + b^2 = c(x + y) $ | Distributive property |
| $ a^2 + b^2 = c^2 $ | Substitution property (since $ x + y = c $) |

statement	reason
\\( \frac{a}{c} = \square, \frac{b}{c} = \square \\)	corresponding sides of similar triangles are proportional.
\\( a^2 = cx, b^2 = cy \\)
\\( a^2 + b^2 = cx + cy \\)
\\( a^2 + b^2 = c(x + y) \\)	distributive property
\\( a^2 + b^2 = c^2 \\)

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QUESTION IMAGE

Question

Explanation:

Step1: Analyze the first proportion

Answer:

Statement	Reason
\( \frac{a}{c} = \frac{x}{a}, \frac{b}{c} = \frac{y}{b} \)	Corresponding sides of similar triangles are proportional.
\( a^2 = cx, b^2 = cy \)	Multiplication property
\( a^2 + b^2 = cx + cy \)	Addition property
\( a^2 + b^2 = c(x + y) \)	Distributive property
\( a^2 + b^2 = c^2 \)	Substitution property (since \( x + y = c \))