Question

consider the diagram and the paragraph proof below
given: right \\( \triangle a b c \\) as shown where \\( c d \\) is an altitude
of the triangle
prove: \\( a^{2}+b^{2}=c^{2} \\)
which is the last sentence of the proof?
\\( \bigcirc \\) because \\( t+e = 1, a^{2}+b^{2}=c^{2} \\)
\\( \bigcirc \\) because \\( t+e = c, a^{2}+b^{2}=c^{2} \\)
\\( \bigcirc \\) because \\( a^{2}+b^{2}=c^{2}, t+e = c \\)
\\( \bigcirc \\) because \\( a^{2}+b^{2}=c^{2}, t+e = 1 \\)
because \\( \triangle a b c \\) and \\( \triangle c b d \\) both have a right angle
and the same angle \\( b \\) is in both triangles, the triangles
must be similar by aa. likewise, \\( \triangle a b c \\) and \\( \triangle a c d \\)
both have a right angle and the same angle \\( a \\) is in
both triangles, so they also must be similar by aa. the
proportions \\( \frac{c}{t}=\frac{a}{t} \\) and \\( \frac{c}{e}=\frac{b}{e} \\) are true because they

Explanation:

Step1: Use similarity of triangles

Since \(\triangle ABC\sim\triangle CBD\), we have \(\frac{a}{f}=\frac{c}{a}\), so \(a^{2}=cf\). Since \(\triangle ABC\sim\triangle ACD\), we have \(\frac{b}{e}=\frac{c}{b}\), so \(b^{2}=ce\)

Step2: Add the two equations

\(a^{2}+b^{2}=cf + ce=c(f + e)\)

Step3: Use the fact that \(f + e=c\)

Substitute \(f + e=c\) into \(a^{2}+b^{2}=c(f + e)\), we get \(a^{2}+b^{2}=c\times c=c^{2}\)

Answer:

Because \( f + e = c\), \( a^{2}+b^{2}=c^{2}\)