in the figure, $\overline{cd}$ is the altitude of right triangle $abc$.
how can similarity be used to prove the pythagorean theorem? move the reasons to the table to complete the proof.
| statement | reason |
| --- | --- |
| $\angle a \cong \angle a$ and $\angle b \cong \angle b$ | |
| $\angle acb \cong \angle adc \cong \angle cdb$ | |
| $\triangle abc \sim \triangle acd$ and $\triangle abc \sim \triangle cbd$ | |
| $\frac{x}{b} = \frac{b}{c}$ and $\frac{y}{a} = \frac{a}{c}$ | |
| $b^2 = cx$ and $a^2 = cy$ | multiplication |
| $a^2 + b^2 = cx + cy$ | |
| $a^2 + b^2 = c^2$ | factoring and substitution |

Question

in the figure, $\overline{cd}$ is the altitude of right triangle $abc$.
how can similarity be used to prove the pythagorean theorem? move the reasons to the table to complete the proof.
| statement | reason |
| --- | --- |
| $\angle a \cong \angle a$ and $\angle b \cong \angle b$ |  |
| $\angle acb \cong \angle adc \cong \angle cdb$ |  |
| $\triangle abc \sim \triangle acd$ and $\triangle abc \sim \triangle cbd$ |  |
| $\frac{x}{b} = \frac{b}{c}$ and $\frac{y}{a} = \frac{a}{c}$ |  |
| $b^2 = cx$ and $a^2 = cy$ | multiplication |
| $a^2 + b^2 = cx + cy$ |  |
| $a^2 + b^2 = c^2$ | factoring and substitution |

Accepted Answer

# Brief Explanations:
1. For $\angle A \cong \angle A$ and $\angle B \cong \angle B$, this is the reflexive property of congruent angles (an angle is congruent to itself).
2. $\angle ACB \cong \angle ADC \cong \angle CDB$ because all are right angles (given $	riangle ABC$ is right-angled at $C$, and $CD$ is an altitude, so these angles equal $90^\circ$).
3. $	riangle ABC \sim 	riangle ACD$ and $	riangle ABC \sim 	riangle CBD$ follows from the AA (Angle-Angle) similarity criterion: we have two pairs of congruent angles for each triangle pair.
4. $\frac{x}{b} = \frac{b}{c}$ and $\frac{y}{a} = \frac{a}{c}$ comes from the property of similar triangles: corresponding sides of similar triangles are proportional.
5. $a^2 + b^2 = cx + cy$ is obtained by adding the two equations $b^2 = cx$ and $a^2 = cy$ together.

# Answer:
| Statement | Reason |
|-----------|--------|
| $\angle A \cong \angle A$ and $\angle B \cong \angle B$ | Reflexive Property |
| $\angle ACB \cong \angle ADC \cong \angle CDB$ | All are right angles |
| $	riangle ABC \sim 	riangle ACD$ and $	riangle ABC \sim 	riangle CBD$ | AA Similarity Criterion |
| $\frac{x}{b} = \frac{b}{c}$ and $\frac{y}{a} = \frac{a}{c}$ | Proportional sides of similar triangles |
| $b^2 = cx$ and $a^2 = cy$ | Multiplication |
| $a^2 + b^2 = cx + cy$ | Addition of equations |
| $a^2 + b^2 = c^2$ | Factoring and substitution |

statement	reason
\\(\angle acb \cong \angle adc \cong \angle cdb\\)
\\(\triangle abc \sim \triangle acd\\) and \\(\triangle abc \sim \triangle cbd\\)
\\(\frac{x}{b} = \frac{b}{c}\\) and \\(\frac{y}{a} = \frac{a}{c}\\)
\\(b^2 = cx\\) and \\(a^2 = cy\\)	multiplication
\\(a^2 + b^2 = cx + cy\\)
\\(a^2 + b^2 = c^2\\)	factoring and substitution

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QUESTION IMAGE

Question

Explanation:

Answer:

Statement	Reason
$\angle ACB \cong \angle ADC \cong \angle CDB$	All are right angles
$\triangle ABC \sim \triangle ACD$ and $\triangle ABC \sim \triangle CBD$	AA Similarity Criterion
$\frac{x}{b} = \frac{b}{c}$ and $\frac{y}{a} = \frac{a}{c}$	Proportional sides of similar triangles
$b^2 = cx$ and $a^2 = cy$	Multiplication
$a^2 + b^2 = cx + cy$	Addition of equations
$a^2 + b^2 = c^2$	Factoring and substitution