Question

Given: ΔABC is a right triangle.
Prove: a2 + b2 = c2

Right triangle BCA with sides of length a, b, and c. Perpendicular CD forms right triangles BDC and CDA. CD measures h units, BD measures y units, DA measures x units.

The two-column proof with missing justifications proves the Pythagorean Theorem using similar triangles:

Which is not a justification for the proof?


Substitution

Addition Property of Equality

Transitive Property of Equality

Distributive Property of Equality

Answer 1

@Michele_Laino

Answer 2

please, we have to apply the similitude criterions

Answer 3

Statement

Justification

Draw an altitude from point C to Line segment AB
Let segment BC = a
segment CA = b
segment AB = c
segment CD = h
segment DB = x
segment AD = y
y + x = c
c over a equals a over y and c over b equals b over x
a2 = cy; b2 = cx
a2 + b2 = cy + b2
a2 + b2 = cy + cx
a2 + b2 = c(y + x)
a2 + b2 = c(c)
a2 + b2 = c2

Answer 4

ok! I got
since triangle BDC and triangle ADC are similar to triangle ABC, we can write:
\[\frac{ a }{ y }=\frac{ c }{ a },\]
and
\[\frac{ b }{ x }=\frac{ c }{ b }\]
from those equation, we can write:
\[a ^{2}=cy, b ^{2}=cx\]
so:
\[a ^{2}+b ^{2}=cx+cy=c(x+y)=c ^{2}\]

Answer 5

nvm I got it right it is transitive property of equality. thanks though

Answer 6

so I've used the first option, the second option, but not the third option

Answer 7

sorry I think that transitive property is used, because if a^2+b^2=c(x+y), and x+y=c, I can write:
a^2+b^2=c^2

Answer 8

it was transitive property I got it right. thanks for your help an explaining I really appreciate it!!(:

Answer 9

thank you!