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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.
Area of triangle C = sum of areas of A and B. All three right triangles are similar, so all three areas are proportional to the side bordering the centre triangle. Hence, α(a2 + b2) = α c2, or dividing by α, we have Pythagoras' theorem.
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English: 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. Thus triangle AHC is similar to triangle ACB by AA test.
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 ...
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