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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
Which theorem is used to prove that aas triangle congruence postulate theorem?
AAS: If Two angles and a side opposite to one of these sides is congruent to thecorresponding angles and corresponding side, then the triangles are congruent.How Do I know? Taking Geometry right now. :)
What are the theorems and postulates you can use to prove triangles are congruent?
Pythagorean's Theorem is one of the most famous ones. It says that the two squared sides of a right triangle equal the squared side of the hypotenuse. In other words, a2 + b2 = c2
Show that if diagonals of a quadrilateral bisects each other then it is a rhombus?
This cannot be proven, because it is not generally true. If the diagonals of a quadrilateral bisect each other, then it is a parallelogram. And conversely, the diagonals of any parallelogram bisect each other. However not every parallelogram is a rhombus.However, if the diagonals are perpendicular bisectors, then we have a rhombus.Consider quadrilateral ABCD, with diagonals intersecting at X, whereAC and BD are perpendicular;AX=XC;BX=XD.Then angles AXB, BXC, CXD, DXA are all right angles and are congruent.By the ASA theorem, triangles AXB, BXC, CXD and DXA are all congruent.This means that AB=BC=CD=DA.Since the sides of the quadrilateral ABCD are congruent, it is a rhombus.
How do you prove that the diagonals and either base of an isosceles trapezoid form an isosceles triangle?
Consider the isosceles trapezium ABCD (going clockwise from top left) with AB parallel to CD. And let the diagonals intersect at O Since it is isosceles, AD = BC and <ADC = <BCD (the angles at the base BC). Now consider triangles ADC and BCD. AD = BC The side BC is common and the included angles are equal. So the two triangles are congruent. and therefore <ACD = <BDC Then, in triangle ODC, <OCD (=<ACD = <BDC) = <ODC ie ODC is an isosceles triangle. The triangle formed at the other base can be proven similarly, or by the fact that, because AB CD and the diagonals act as transversals, you have equal alternate angles.
