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English: Simplified version of similar triangles proof for Pythagoras' theorem. In triangle ACB, angle ACB is the right angle. CH is a perpendicular on hypotenuse AB of triangle ACB. In triangle AHC and triangle ACB, ∠AHC=∠ACB as each is a right angle. ∠HAC=∠CAB as they are common angles at vertex A.
The apparent triangles formed from the figures are 13 units wide and 5 units tall, so it appears that the area should be S = 13×5 / 2 = 32.5 units. However, the blue triangle has a ratio of 5:2 (=2.5), while the red triangle has the ratio 8:3 (≈2.667), so the apparent combined hypotenuse in each figure is actually bent. With the bent ...
There are several elementary results concerning similar triangles in Euclidean geometry: [9] Any two equilateral triangles are similar. Two triangles, both similar to a third triangle, are similar to each other (transitivity of similarity of triangles). Corresponding altitudes of similar triangles have the same ratio as the corresponding sides.
The two triangles on the left are congruent. The third is similar to them. The last triangle is neither congruent nor similar to any of the others. Congruence permits alteration of some properties, such as location and orientation, but leaves others unchanged, like distances and angles. The unchanged properties are called invariants.
If the three triangles XBC, YCA, ZAB, constructed on the sides of the given triangle ABC as bases, are similar, isosceles and similarly situated, then the lines AX, BY, CZ concur at a point N. If the common base angle is θ, then the vertices of the three triangles have the following trilinear coordinates.
A similar proof uses four copies of a right triangle with sides a, b and c, arranged inside a square with side c as in the top half of the diagram. [6] The triangles are similar with area , while the small square has side b − a and area (b − a) 2. The area of the large square is therefore