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Question
given: △abc is a right triangle and cd⊥ab. prove: ac² + bc² = ab². proof: 1. ∠acb≅∠adc≅∠cdb 2. ∠cab≅∠dac ∠abc≅∠cbd 3. △abc - △acd △abc - △cbd 4. $\frac{ac}{ab}=\frac{ad}{ac}$ and $\frac{bc}{ab}=\frac{db}{bc}$ 5. ac²=(ab)(ad) bc²=(ab)(db) 6. ac² + bc²=(ab)(ad)+(ab)(db) 7. ac² + bc² = ab(ad + db) reasons: angles with the same measure are congruent, reflexive property, aa similarity criteria,?, cross - multiplication, addition, distributive property
Step1: Recall similarity - triangle properties
From $\frac{AC}{AB}=\frac{AD}{AC}$ and $\frac{BC}{AB}=\frac{DB}{BC}$, by cross - multiplying we get the equations in statement 5. Cross - multiplying is a valid operation when dealing with proportion equations $\frac{a}{b}=\frac{c}{d}$ which gives $ad = bc$.
Step2: Analyze the addition step
In step 6, we add the two equations $AC^{2}=(AB)(AD)$ and $BC^{2}=(AB)(DB)$ from step 5. According to the addition property of equality, if $a = b$ and $c = d$, then $a + c=b + d$.
Step3: Apply distributive property
In step 7, we factor out $AB$ from $(AB)(AD)+(AB)(DB)$ using the distributive property $ab+ac=a(b + c)$. Since $AD + DB=AB$ (by the segment - addition postulate where point $D$ lies on segment $AB$), we get $AC^{2}+BC^{2}=AB^{2}$.
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The missing reason in step 5 is Cross - multiplication.