[alogo] 1. Projecting on tangents

Draw the tangents {AB,AC} to circle c from point A. For every point D on the circle holds the basic relation:
DE2 = DB'*DC',
E denoting the projection on the chord BC of contact points and {B',C'} denoting correspondingly the projections on the tangents {AB,AC}.

Follows immediately from the similarity of triangles DBC, DB'E, DEC'.
Notice that quadrangles DEBB' and DC'CE are similar. If triangle DEC' is modified so that it remains similar to itself while C' glides on AC then E will move on BC (see Similarly_Rotating.html ). Analogous statement holds for triangle DB'E.


[alogo] 2. Products of distances

For every point D of the circumcircle of a triangle the products of distances are equal:
DA'*DB'*DC' = DA''*DB''*DC''.
Here {A',B',C'} are the projections of D on the sides of ABC and {A'',B'',C''} the projections of D on the tangents at {A,B,C} correspondingly.

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It suffices to apply the basic relation of (1) for each ange of the tangential triangle: DB'2 = DA''*DC'', DC'2 = DB''*DA'', DA'2 = DC''*DB''. Then multiply the equations and simplify.

Remark The same proof works for the generalization to n-sided cyclic polygons A1...An and the corresponding tangential polygon B1...Bn, created by drawing tangents at A1,...,An. If D is a point on the circumcircle of the polygon and {X1,...,Xn} denotes the projections of D on the sides and {Y1,...,Yn} its projections on the sides of the tangential polygon then
DX1* ... *DXn = DY1* ... *DYn.

[alogo] 3. Projecting from the circumcircle

Let P be a point on the circumcircle c(O,r) of a cyclic quadangle ABCD.
[1] Project P on two consecutive sides, {AB,AD} say, to create triangle PGH. Analogously project on the other two consecutive sides {CB,CD} to create triangle PEF. The two triangles are similar. This follows easily by angle chasing argument.
[2] The circumcenters {OA,OC} of triangles PGH and PEF have OAOC parallel to AC and half its length.
[3] Analogous statements hold for the triangles created similarly by selecting the other pair of opposite vertices {B,D}.
[4] The circumcenters of the four circumcircles {OA,OB,OC,OD}, the center O and P lie all on a circle c' with radius half the radius r of c, hence also tangent to c.
[5] Project D on pairs of opposite sides {AB,CD} to create triangle PJI and {AD,BC} to create PKL.
area(PLK)/area(PJI) = sin(x)/sin(y),
where {x,y} are the angles formed by opposite sides. Thus this area-quotient is independent of the position of P on the circle.

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[2,3,4] Follows from the fact that {PA,PB,PC,PD} are diameters of the circumcircles of triangles considered.
[5] Is a consequence of the previous similarities.

Remark Quadrangle OAOBOCOD is homothetic from P to ABCD, the hothety ratio being 1/2.

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