Let D and E be points on the hypotenuse AB such that BD = BC and AE = AC. Let AD = x, DE = y, BE = z. Then AC = x + y, BC = y + z, AB = x + y + z. The Pythagorean theorem is then equivalent to the algebraic identity
(y + z)² + (x + y)² = (x + y + z)².
Which simplifies to
y² = 2xz.
To see that the latter is true calculate the power of point A with respect to circle B(C), i.e. the circle centered at B and passing through C, in two ways: first, as the square of the tangent AC and then as the product AD·AL:
(x + y)² = x(x + 2(y + z)),
which also simplifies to y² = 2xz.
This is algebraic proof 101 from Loomis' collection. Its dynamic version is available separately.
Let D and E be points on the hypotenuse AB such that BD = BC and AE = AC. Let AD = x, DE = y, BE = z. Then AC = x + y, BC = y + z, AB = x + y + z. The Pythagorean theorem is then equivalent to the algebraic identity(y + z)² + (x + y)² = (x + y + z)².Which simplifies toy² = 2xz.To see that the latter is true calculate the power of point A with respect to circle B(C), i.e. the circle centered at B and passing through C, in two ways: first, as the square of the tangent AC and then as the product AD·AL:(x + y)² = x(x + 2(y + z)),which also simplifies to y² = 2xz.This is algebraic proof 101 from Loomis' collection. Its dynamic version is available separately.
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