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

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- #1

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

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TD

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Proof of what exactly?

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

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TD

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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.

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wouldnt the proof be something like this

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

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

- #6

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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.mathmike said:wouldnt the proof be something like this

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

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Tide

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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.

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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.

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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