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March 29th, 2017, 11:22 PM   #1
Joined: May 2016
From: new zealand

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Please answer quick. Question on similar triangles

My teachers keeps telling me that I'm using a method that should not work for all similar triangle questions but it does. The differences of the same side on the two different sized but same angled triangles should be proportionate right?
So if the difference is the same, the sides should be the same. To keep two different lengths in proportion while making them bigger, should the differences be in proportion. Example, if one side is 15 and the other proportionate side is 5, and you are adding 3 to 15, you cannot add 3 to 5 and still keep the sides proportionate. You have to add a third of what you add to 15 because 15 is three times larger than 5. So if we add 1 to 5 and 3 to 15, the sides should still be proportionate. Am I wrong and how do I explain this to my teacher?
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March 30th, 2017, 09:52 PM   #2
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Until you gave your example, I had no idea what you were talking about. Your explanation is obscure to say the least. But if what you are saying is

$\dfrac{x}{y} = c\ and\ z \ne -\ y \implies \dfrac{x + cz}{y + z} = c$,

that is true.

$Given\ \dfrac{x}{y} = c\ and\ z \ne -\ y.$

$z \ne -\ y \implies y + z \ne 0.$

$And\ \dfrac{x}{y} = c \implies x = cy \implies x + cz = cy + cz = c(y +z).$

$\therefore \dfrac{x + cz}{y + z} = \dfrac{c(y + z)}{(y + z)} = c.$

Last edited by JeffM1; March 30th, 2017 at 09:54 PM.
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March 31st, 2017, 06:38 PM   #3
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Do you mean ASS similarity? Well, this situation might be correct, but sometimes two different triangles will fit in this rule at the same time.

If the angle is equal to 90º(which defined as RHS, or H.L), or bigger than 90º, your opinion will work.

Hope this will help you.
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