The general solution would be the first triangle having sides of a, ab and ab² and the second having sides of ab, ab² and ab³ (a>0, b ≠0 or 1) .
In ganesh's example, a = 1 and b = √2.
]]>A 4 con triangle is very very possible... even in 30/60/90's... just try it.
The thing about the triangle is that the equal sides/angles dont have to be corressponding, which is why the SAS AAS etc will not prove it to be congruent...
]]>A 5-con triangle would be a non-congruent triangle with 5 similar properties. For this to happen, it would need at least 2 similar angles and 2 similar sides, and this satisfies the AAS rule, so it would have to be congruent and so be a 6-con triangle.
In fact, I don't think you can even have 4-con triangles.
]]>What 5-Con triangles are, are any two triangles who have 5 (not necessarily corresponding) sides and angles equal.
Triangles obviously have 3 angles and 3 sides, so 5 of the 6 must be equal.
I know the 3 angles must be the same and 2 sides equal, because if all 3 sides are equal lengths, all 3 angles must be also.
So pretty much i need to find 2 triangles, who have all three angles equal, but only 2 sides equal. Any ideas?
(Ive been working with 30/60/90 triangles but cant seem to come up with it)
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