## Τρίτη, 26 Φεβρουαρίου 2013

### CIRCLES OF QUADRILATERAL

PROBLEM:

Let ABCD be a quadrilateral. To construct two EQUAL and TANGENT circles (K),(L) such that: (K) touches the sides of angleA, and (L) the sides of angleC of ABCD.

Analysis:

K lies on the bisector of angA, and L on the bisector of angC (K,L: the centers of (K),and (L), resp.) Let O be the intersection point of these bisectors. Draw KQ _|_ AB, LI _|_ BC and OZ _|_ AB, OE _|_ BC (Q, Z on AB; I,E on BC).

We have: KQ = LI, and KL = KQ + LI = 2KQ = 2LI (1)

AQK ~ AZO ==> KQ/AK = OZ/OA (2)

Draw Ox // KL, and let M be the intersection point of Ox,AL.

AKL ~ AOM ==> KA/KL = OA/OM (3)

[(1) and (3)] ==> KA/2KQ = OA/OM (4)

[(2) and (4)] ==> OA/2OZ = OA/OM ==> OM = 2OZ.

Draw My // AC, and let H be the intersection point of My and CO.

CLI ~ COE ==> CL/LI = CO/OE (5)

AC // MH ==> CL/CH = AL/AM = KL/OM (6)

(5) ==> CL = LI*OC/OE (7)

(6) and (7) ==> LI* OC/OE / CH = KL/OM (and since KL = 2LI) ==> OC/2OE = CH/OM ==> (since OM = 2OZ) 2OE/OC = 2OZ/CH ==> OE/OC = OZ/CH, that is CH is the 4th proportional of OE,OC,OZ.

Synthesis:

Let O be the intersection point of bisector angA and bisector of angC. Draw OZ _|_ AB, OE _|_ BC. Draw the circle (O, 2OZ). Take on OC the segment CH such that CH = the 4th proportional of OE,OC,OZ. Draw Hy//AC, which intersects the circle (O, 2OZ) at M near to C. Draw MA, intersecting OC at L. Draw Lx//OM, intersecting OA at K. The points K and L are the centers of the circles in question.

Proof: For the reader

Bibliography:

ΠΑΡΑΡΤΗΜΑ του ΔΕΛΤΙΟΥ της ΕΛΛΗΝΙΚΗΣ ΜΑΘΗΜΑΤΙΚΗΣ ΕΤΑΙΡΕΙΑΣ [=Supplement of the Bulletin of the Greek Mathematical Society], March 1967, pp. 160 - 161