- Dear Antreas and Alexey,

[AZ]> Dear colleagues!

Here is a proof:

> Given circumscribed quadrilateral ABCD. I is its incenter,

> M, N are

> the midpoints of AC and BD.

> Then IM/IN=AC/BD iff ABCD is cyclic.

If IA = a, IB = b, IC = c, ID = d

angle x = (A+C)/2 and

r is the radious of incircle ABCD then

a = r/sin(A/2), b = r/sin(B/2)

c = r/sin(C/2), d = r/sin(D/2)

2.sin(A/2).sin(C/2) = cos(x-C) - cos(x)

2.sin(B/2).sin(D/2) = cos(x) - cos(x+D)

sin²(A/2)+sin²(C/2)=(1-cosA+1-cosC)/2

=1-cos(x).cos(x-C)= P

sin²(B/2)+sin²(D/2)=1-cos(x).cos(x+D) = Q

4.IM² = 2(a²+c²)-AC²

4.IN² = 2(b²+d²)-BD²

Since

AC² = a²+c²-2ac.cos(360-D-A/2-C/2) or

AC² = a²+c²-2ac.cos(x+D)

BD² = b²+d²-2bd.cos(360-C-B/2-D/2) or

BD² = b²+d²+2bd.cos(x-C)

applying ratio properties we get

IM²/IN² = AC²/BD² <=>

AC²/BD² = (a²+c²)/(b²+d²) <=>

(a²+c²)/(b²+d²)= -ac.cos(x+D)/bd.cos(x-C) <=>

[sin²(A/2)+sin²(C/2)]/[sin²(B/2)+sin²(D/2)]=

=-sin(A/2)sin(C/2)cos(x+D)/sin(B/2)sin(D/2)cos(x-C) <=>

[1-cos(x).cos(x-C)]/[1-cos(x).cos(x+D)]=

[cos(x)-cos(x-C)]cos(x+D)/[cos(x)-cos(x+D)]cos(x-C) <=>

P/Q = R/S = (P-R)/(Q-S) = 1 <=> P = Q (1)

because

P-R = Q-S = 1-cos(x)[cos(x-C)+cos(x+D)]+cos(x-C)cos(x+D)

and P - R =/= 0 because

P-R > 1-[cos(x-C)+cos(x+D)]+cos(x-C)cos)x+D) or

P-R > [1-cos(x)][1-cos(x+D)] > 0

Hence from (1) we get

P = Q <=> 1-cos(x).cos(x-C) = 1-cos(x).cos(x+D) <=>

cos(x)[cos(x-C)-cos(x+D)] = 0 <=>

2cos(x).sin((C+D)/2).sin((A+D)/2) <=>

cos(x) = 0 <=> (A+C)/2 = 90 <=> A+C = 180

or ABCD is cyclic.

Is there a simpler proof?

Best regards

Nikos Dergiades

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μηνυμάτων http://login.yahoo.com/config/mail?.intl=gr > Dear Vladimir,

[VD]

> > [APH]

> >> Let P be a point inside parallelogram ABCD,

> >> such that angles ABP = ADP.

> >> To prove that angles PAB = PCB

>

> > [ND]

> > If P is an arbitrary point then the reflexion D' of D

> > in PA and the reflexion C' of C in PB are equidistant

> > from the circumcenter O of PAB. i.e. OC' = OD'.

> > (See the following lemma)

> > Hence in case of /_ABP = /_ADP = /_AD'P the point

> > D' lies on the circumcircle of PAB and by the lemma

> > C' lies on the circumcircle and we get

> > /_PAB = /_PC'B = /_PCB.

> I think the shortest, at least the most elementary,

Yes. You are right.

> solution is to notice that

> if Q is the translate of P by vector AB=CD, then both

> equations for angles

> are equivalent to the cyclicity of the quad BPCQ

> (<BCQ=<ADP=<ABP=<BPQ

> and similarly for the second equation).

Very nice proof!

[ND]> > LEMMA. If ABC is a triangle with circumcenter O and

[VD]

> > BB' = CC' are two vectors then the reflexion

> > B" of B' in AB and the relexion C" of C' in AC

> > are equidistant from O with angle B"OC" = 2.A

> >

> > I have a nice proof but I have no time to write it

> > and since I am leaving home to Mytilene I will see

> > your clever comments I hope after three days

> > from an Internet cafe.

> Nikos, I wonder if your proof is based on the fact that

Not exactly.

> the

> composition of two reflections

> is a rotation by twice the angle between their axes.

If O is the circumcenter of ABC and

Ob is the reflexion of O in AB and

Oc is the reflexion of O in AC then

since for vectors BB' = CC', BOb= OA = COc

the triangles BB'Ob, CC'Oc are congruent

and directly similar,

the triangles BB'Ob, BB"O are congruent

and inversely similar,

the triangles CC'Oc, CC"O are congruent

and inversely similar, we conclude that

the triangles BB"O, CC"O are congruent

and directly similar.

Since the rotation about O by an angle 2.A

moves B on C, B" must move on C".

Hence PB" = OC" and angle B"OC" = 2.A

Best regards

Nikos

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