![]() As a result, the area of △BDF is twice the area of △DFE. So △ABF and △DEF are similar triangles and AB/ED=BF/FE.īut the ratio AB/ED is the same as BC/BE, because △ABC and △BDE are similar triangles with a scale factor of 1:2, so AB/ED=2/1=BF/FE, and we have shown that BF=2FE. ![]() ∠AFB ≅ ∠DFE because they are vertical angles. m∠ABF=m∠DEF=60°, since △ABC and △BDE are both equilateral triangles. Now let's look at two triangles where FE and BF are sides - △ABF and △DEF. △ABC and △BDE are both equilateral triangles, so they are similar triangles, and the scale factor is 1:2. To show that BF=2FE, we will use similar triangles, and the hint we have is that we already have two line segments with that ratio: BE=EC, so BC=2BE. Also the Pythagorean theorem can be used for non. For two triangles to be similar the angles in one triangle must have same value as the angles in another triangle. If another triangle can be divided into two right triangles (see Triangle), then the area of the triangle may be able to be determined from the sum of the two constituent right triangles. Here, the triangles whose areas we are asked to compare (△BDF and △DFE) have the same height, DG, so their areas will be proportional to the length of their bases, BF and FE. All isosceles right triangles are similar since corresponding angles in isosceles right triangles are equal. In a previous similar problem, we noted that problems that require finding the ratio of areas often rely on using similar triangles, or on triangles that have the same base or same height. Show that the area of △BDF is twice the area of △DFE. △ABC and △BDE are both equilateral triangles. We will use this in the following geometry problem. ![]() So by definition, all regular polygons with the same number of sides are similar to each other.Īn equilateral triangle is a regular polygon with 3 sides, so all equilateral triangles are similar. A regular polygon is one in which all angles are equal and all sides are equal. ![]()
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