An equilateral triangle has 3 equal angles and 3 equal sides. The angles add up to 1800 and there are 3 equal angles so each is 600
an equilateral triangle
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In order for a triangle to be congruent the two triangles have to be the same shape and size, thus they are congruent if they can be moved into an isometry or any other combination. But you're asking how a question which has two possibilities. Assuming that you have two triangles whose sides are equivalent which makes the areas equal to each other then you can state the side-side-side rule which is if the three sides of one triangle is equivalent to the other three sides of the other triangle then they are congruent. But if you have an angle present in the triangles you could argue the angle angle side rule, but if the angles are joint you would argue the angle side angle. But if one triangle has one degree and the other one has a different degree then they will not be congruent.
No. You can know all three angles of both and all you can say is that the triangles are similar. Or with any pair of congruent sides you can have an acute angle between them or an obtuse angle.
If of triangle ABC and A'B'C' sides AB = A'B' and AC = A'C', and the included angle at A is larger than the included angle at A*, then BC > B'C'.Proof:A A'/|\ /|/ | \ / |/ | \ / |/ | \ B'/ |B | X \C |C'DWe construct AD such that AD = A'C' = AC and angle BAD = angle B'A'C'.Triangles ABD and A'B'C' are congruent. Therefore BD = B'C'.Let X be the point where the angle bisector of angle DAC meets BC.From the congruent triangles AXC and AXD (SAS) we have that XD = XC.Now, by the triangle inequality we have that BX + XD > BD, so BX + XC > BD, and consequently BC > BD = B'C'.
There is no AAA theorem since it is not true. SSS is, in fact a theorem, not a postulate. It states that if the three sides of one triangle are equal in magnitude to the corresponding three sides of another triangle, then the two triangles are congruent.
non square rhombus