Consider a circle c(J,r) and a point A at which rotates a right angle intersecting the circle at points {B,C,D,E}. Show that the sum of squares (0) AB2 + AC2 + AD2 + AE2 = 4r2, is a constant independent from the location of the point A [Emch p. 115].
[1] There is a rectangle FGHI with vertices at the middles of the sides of the cyclic and orthodiagonal quadrangle BECD (see Orthodiagonal.html ). [2] The similarity of triangles ACE and ADB implies that AH is orthogonal to BD, hence A is the Mathot point (or anticenter) of BECD (see Mathot.html ), coinciding with the intersection point of the diagonals of BECD (which is orthodiagonal). [3] This implies that AFJH is a parallelogram and the center K of the rectangle FGHI is the middle of JA. [4] The opposite sides CE and DB of BECD intersect at M on the pola LM of A. [5] FJHM is cyclic having opposite right angles the intersection L of AJ with the polar of A is on its circumcircle. Thus the radius R2 = HK2 = JK*KL. Setting JA = d, and taking into account that JA*JL = r2, we obtain JL = r2/d => R2 = (d/2)*(r2/d-d/2) = r2/2 - d2/4, which implies: 4R2 = 2r2 - d2 (*). [6] The sum of the squares S = AC2+AB2+AD2+AE2 = AB2 + DE2 - 4(r+d)(r-d). But AB2 + DE2 = 4(HG2+GF2) = 16R2 => S = 16R2 - 4(r2-d2), which by (*) gives: S = 4r2.
The sides of the cyclic quadrangle BECD created by the right angle rotating about A evelope a conic. In the case point A is inside the circle c(J,r) this is an ellipse with foci at J and A and axis 2R, where R is the radius of the circle enclosing the rectangle FGHI (see (*) of previous section).
Though the property is a consequence of a more general one (see Conic_Intersect_Variable_Angle.html ), the direct proof follows immediately from the preceding section by observing a triangle equality. If namely J' is the reflected of J on CE, then by the discussion in the previous paragraph follows that triangles J'HA and HJF are equal, hence AJ' = HF = 2R.
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