Question

given: δabc is a right triangle. prove: ( a^2 + b^2 = c^2 ) the following two - column proof with missing justifications proves the pythagorean theorem using similar triangles: statement justification draw an altitude from point c to ( overline{ab} ) let ( overline{bc} = a ) ( overline{ca} = b ) ( overline{ab} = c ) ( overline{cd} = h ) ( overline{db} = y ) ( overline{ad} = x ) ( y + x = c ) ( \frac{c}{a}=\frac{a}{y},\frac{c}{b}=\frac{b}{x} ) ( a^2 = cy, b^2 = cx ) ( a^2 + b^2 = cy + b^2 ) ( a^2 + b^2 = cy + cx )

Explanation:

Step1: Similar triangles proportion

In right - triangle \(ABC\) with altitude \(CD\), \(\triangle ABC\sim\triangle CBD\) and \(\triangle ABC\sim\triangle ACD\).
For \(\triangle ABC\sim\triangle CBD\), we have \(\frac{c}{a}=\frac{a}{y}\) (by the property of similar triangles, corresponding side ratios are equal). Cross - multiplying gives \(a^{2}=cy\).
For \(\triangle ABC\sim\triangle ACD\), we have \(\frac{c}{b}=\frac{b}{x}\). Cross - multiplying gives \(b^{2}=cx\).

Step2: Substitute \(b^{2}\)

Substitute \(b^{2}=cx\) into \(a^{2}+b^{2}=cy + b^{2}\), we get \(a^{2}+b^{2}=cy+cx\).

Step3: Factor out \(c\)

Factor out \(c\) from the right - hand side: \(a^{2}+b^{2}=c(y + x)\).
Since \(y + x=c\) (given \(y+x = c\) from the segment addition postulate on \(AB\)), then \(a^{2}+b^{2}=c\times c=c^{2}\).

Answer:

The proof shows that \(a^{2}+b^{2}=c^{2}\) using the properties of similar triangles and algebraic substitution.