## 1. Rotating right angle

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
[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.

## 2. Tangent conic

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.