## 1. Four circles related to a trapezium

Consider a trapezium ABCD and the intersection point E of its diagonals.
1) The circumcircles of triangles ABE, DEC constructed on the parallel sides are tangent.
2) The circumcircles of triangles AED, BEC on the non-parallel sides intersect at a second      point F, such that the two triangles FAD and FCB are similar.
3) The centers {G, J, I, H} of the circumcircles of the four triangles considered above are      vertices of a parallelogram.
4) Point F is the center of a similarity turning the non-parallel side AD to CB.
5) If K is the intersection of AD and BC, then quadrangle KDFB is cyclic.
6) F is the second intersection point of the circumcircles of triangles KAC and KDB.

(1) Follows from the equality of angles DCE and EAB which measure the inclination of the tangent to DB.
(2) First show that FDB is similar to FAC. In fact, angles FAE and FDE are equal, since A, D are on the same circle and view the same chord of it EF. Analogously angles FBE and FCE are equal.
Thus, FDB and FAC are similar triangles and working at F we see that FB/FC = FD/FA and the angles there of triangles FAD and FCB are equal hence the proof of the claim.
(3) The angle HJI is equal to FEB, the two angles having corresponding sides orthogonal.
Analogously angles JHG and FED are complementary. It follows that JI and HG are parallel.
Analogously sides HI and GJ are parallel proving the claim.
(4) is a consequence of the similarity of triangles FAD and FCB.
(5) follows also from this similarity, since then angle FBC is equal to the external opposite angle ADF of the quadrilateral KDFB.
(6) follows from (5).

## 2. Two orthogonal lines

Continuing the discussion on the previous figure, reflect the similarity center F on the non-parallel sides to obtain points {G, H}. Let also {M,N} be the middles of {FH, FG} respectively.
(1) Then line GH is orthogonal to KE.
(2) The intersection point L of these two lines, the similarity center F, and points M, N are vertices       of a parallelogram.
(3) L is the orthocenter of triangle KMN.

By the similarity of triangles FAD and FCB proved in section-1, follows that the right-angled triangles FAM and FCN are similar. Thus triangle MFN is similar to the similar triangles FAC and FDB.
Thus the ratio FM/FN = AD/BC. This is the inverse of the ratio of distances of points on KE.
In fact line KE is the median of triangles KDC and/or KAB and is characterized by the property of its points X to have distances from the sides inverse proportional to them:
This implies (see Isogonal.html ) that lines KF and KE are isogonal conjugate and symmetric with respect to the bisector of angle AKB. Consequently angles EKC and FKA are equal and also equal are the angles MKE and FKN. But angle FKN is equal to FMN since FMKN is cyclic quadrilateral.
In total angle EKM is equal to FMN and sides {FM, MK} are orthogonal. Hence the other two sides of the equal angles will be also orthogonal i.e. {MN, KE} are orthogonal. This implies the first claim, since GH is parallel (and double) to MN.
To prove (2) notice that KHG is isosceles, hence L is the middle of HG. Thus LN and FM are equal and parallel. This implies also (3), since then LN being parallel to FM will be also orthogonal to KM.
For additional properties of trapezia see the files Trapezium.html , Trapezium_0.html .