Given: ΔABC is a right triangle. Prove: a2 + b2 = c2 The following two-column proof with missing justifications proves the Pythagorean Theorem using similar triangles: Statement Justification Draw an altitude from point C to Let = a = b = c = h = y = x y + x = c a2 = cy; b2 = cx a2 + b2 = cy + b2 a2 + b2 = cy + cx a2 + b2 = c(y + x) a2 + b2 = c(c) a2 + b2 = c2 Which is not a justification for the proof? Substitution Addition Property of Equality Transitive Property of Equality Distributive Property of Equality

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apocritia apocritia · Answer 1 · 2018-12-26T04:59:22+01:00

Answer:

Transitive property of equality is not a justification for the proof.

Step-by-step explanation:

We draw a right angle ΔACB. CD is perpendicular to AB.

Let AC = a , BC = b , AB = c and CD = h

Now in ΔABC and ΔACD

∠C = ∠D and ∠A = ∠A

from AA similarity postulate

ΔABC similar to ΔACD.

Hence,

[tex]\frac{c}{a}[/tex] = [tex]\frac{a}{x}[/tex]

[tex]a^{2}[/tex] = c × x ·····················(1)

Now in ΔABC and ΔCBD

∠C = ∠D and ∠B = ∠B

from AA similarity postulates

ΔABC similar to ΔCBD

Hence,

[tex]\frac{c}{b} = \frac{b}{y}[/tex]

[tex]b^{2}[/tex] = c × y······················(2)

Add equation (1) and (2)

[tex]a^{2}[/tex] + [tex]b^{2}[/tex] = cx + cy

[tex]a^{2}[/tex] + [tex]b^{2}[/tex] = c(x+y)

[tex]a^{2}[/tex] + [tex]b^{2}[/tex] = [tex]c^{2}[/tex] [because x+y=c]

Transitive property is not useful for this proof.