# Similar triangles, looking for demonstration

#### Werg22

I need an algebric proof of the theorem of similar triangles (C/c=A/a=B/b).

#### TD

Homework Helper
Proof of what exactly?

#### Werg22

TD said:
Proof of what exactly?
For two similar triangles, that A/a=C/c=B/b.

#### TD

Homework Helper
I take it you mean the sides with A,B,C and a,b,c?

Your proof would depend on who you define similar triangles, I assume that's saying that the corresponding angles of both triangles are the same.

#### mathmike

wouldnt the proof be something like this

Sin A / a = Sin B /b = Sin C / c

#### Werg22

mathmike said:
wouldnt the proof be something like this

Sin A / a = Sin B /b = Sin C / c
Well trigonometric functions don't proove the theorem, they are a product of it so it is not really a proof... Anyway I'm putting my mind into it and I hope to find why it is so.

#### Tide

Homework Helper
Werg,

Your problem is actually pretty straightforward but the question I have for you is "What have you tried so far?" This sounds like a homework problem.

#### Werg22

I was finally able to proove it. Here is my solution, but perhaps there is simplest one and I would like someone to show it...

1.Two similar right triangles of sides a,b,c and x,y,z.

The following is true:

(c+z)^2=(a+x)^2+(b+y)^2
c^2 + z^2 + 2cz=a^2 + x^2 + 2xa + b^2 + y^2 + 2by
Since c^2=a^2 + b^2, and z^2=x^2 + y^2,
2cz=2xa + 2by
cz=xa + by
For simplifying purpose, square the equation;

(cz)^2=(xa + by)^2
then
(cz)^2=(xa + by)^2

(x^2 + y^2)(a^2 + b^2)=(xa + by)^2

(ax)^2 + (xb)^2 + (ay)^2 + (by)^2=(ax)^2 + (by)^2 + 2xaby

So

(xb)^2 + (ay)^2 - 2xaby = 0

Factorising, we find that

(xb-ay)^2=0
xb=ay
y/b=x/a

To proove that z/c=x/a=y/b,

x/a=y/b
y=bx/a
and
b=ay/x

(z/c)^2=(x^2 + y^2)/(a^2 + b^2);
(z/c)^2=(x^2 + (bx/a)^2)/(a^2 + (ay/x)^2)
(z/c)^2=(x^2(1+(b/a)^2)/(a^2((y/x)^2 + 1)

Since we know that b/a=y/x, we obtain the result
(z/c)^2=(x/a)^2
z/c=x/a=y/b

Since any triangle can be decomposed into two right triangles, it is easy to prove that the rule is general.

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