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Eight math problems
1 is a parallelogram. It is proved by congruence that triangle DBE and FEC are congruent with triangle ABC respectively. Take one of them as an example. DB=AB, BE=BC (both because they are equilateral triangles), and the angle DBE is equal to the angle ABC (both are 60 degrees minus the included angle EBA). In the same way, it is proved that another triangle makes the triangle DBE and FEC congruent, and because it is an equilateral triangle, DE=CF=AF, DB=DA=EF. So the opposite sides are equal, which is a parallelogram.

2. The four sides of the diamond need to be equal, and the lengths of two adjacent sides are actually the lengths of AB and AC of the triangle (equilateral triangle and congruence can be proved, the same as the first step). So as long as AB=AC, that is, the triangle is an isosceles triangle.