{B,C,D,E}. Show that the sum of squares

(0) AB

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

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 R

we obtain JL = r

4R

[6] The sum of the squares S = AC

AB

S = 4r

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.

Mathot.html

Conic_Intersect_Variable_Angle.html

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