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Two triangles are similar if: 1) 3 angles of 1 triangle are the same as 3 angles of the other or 2) 3 pairs of corresponding sides are in the same ratio or 3) An angle of 1 triangle is the same as the angle of the other triangle and the sides containing these angles are in the same ratio. So if they are both equilateral, then they both have three 60 degree angles since equilateral triangles are equiangular as well. Then number 1 above tell us by AAA, they are similar.
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A diagonal separates the parallelogram into?
Two congruent triangles.. To prove it, use the SSS Postulate.
It appears from the name of the HL Theorem that you actually need to know that only two parts of two triangles are congruent in order to prove two triangles congruent Is this the case?
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.
What is the donkey theorem?
When trying to prove two triangles congruent, you can use SSS, SAS, ASA, AAS, HL, and HA patterns. However, the pattern A S S doesn't work. Instead of spelling or saying this word in class, you can refer to it as "the donkey theorem". You can look at the pattern in the two triangles and say "these two triangles are not congruent because of the donkey theorem." You CANNOT prove triangles incongruent with 'the donkey theorem', nor can you prove them congruent. It's mostly sort of a joke, you could say, but it's never useful. The reason is that if the two triangles ARE congruent, then of course there will be an unincluded congruent angle as well as two congruent sides. The theorem doesn't do anything left, right, forward or backward. It's not even really a theorem. :P
Prove diagonals are equal in a rectangle?
Suppose ABCD is a rectangle.Consider the two triangles ABC and ABDAB = DC (opposite sides of a rectangle)BC is common to both trianglesand angle ABC = 90 deg = angle DCBTherefore, by SAS, the two triangles are congruent and so AC = BD.
How do you prove lines are parallel in similar triangles?
Given:In a triangle ABC in which EF BC To prove that:AE/EB=AF/FC Construction:Draw EX perpendicular AC and FY perpendicular AB Proof:taking the ratios of area of triangle AEF and EBF and second pair of ratio of area of triangle AEF and ECF. We get AE/EB and AF/FC we know that triangle lie b/w sme and same base is equal in area therefore area of EBF I equal to area of ECF therefore AE/EB=AF/FC HENCE PROVED
