Examples of using harmonic divisions and polarity

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pardesi

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pardesi

#1 h Oct 2, 2007, 5:22 PM • 5 Y

Y by thunderz28, Adventure10, and 3 other users

Can anyone please tell me some references or problems about harmonicity i.e. polarity?

[Edited by pohoatza: Title + content of some following messages for maintaining them (the title + next posts) one related to each other.]

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BaBaK Ghalebi

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BaBaK Ghalebi

#2 h Oct 2, 2007, 7:39 PM • 10 Y

Y by Adventure10, Mango247, and 8 other users

I will start with an example of using it in proving the Butterfly theorem. Here it goes:

let $AB$ be an arbitary chord of circle $w$ ,let $P$ be the midpoint of segment $AB$ ,now let $KM$ and $LN$ be two other arbitary chords from $w$ which pass thruogh $P$ ,let $LM,KN$ intersect $AB$ at $X,Y$ respectively,prove that $PX = PY$ .

proof:
note that $L$ is a point on $w$ and we know that $LA,LB,LM,LN$ intersect the circle at $A,B,X,P$ respectively therefore:

$(AB,MN) = (AB,XP)$ (I)
wrt point $L$ .

and similary we get that:
$(AB,MN) = (AB,PY)$
wrt point $K$ .

now let $Y'$ be the reflection of point $Y$ wrt $P$ ,so its sufficient to show that $Y' = X$ ...
the reflection of points $A,B,P,Y$ wrt point $P$ are $B,A,P,Y'$ respectively thus:

$(AB,PY) = (BA,PY') = (AB,Y'P)$ (II)

$(I),(II)\Rightarrow (AB,Y'P) = (AB,XP)$
$\Rightarrow X = Y'$
as wanted...

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BaBaK Ghalebi

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BaBaK Ghalebi

#3 h Oct 2, 2007, 7:54 PM • 5 Y

Y by Adventure10, Mango247, and 3 other users

also with the same method you can prove one of the generalizations of it:

let $AB$ be an arbitary chord of circle $w$ and let $P$ be an arbitary point on $AB$ ,let $LN,KM$ be two other arbitary chords of $w$ which pass thruogh $P$ ,now assume that $LM,KN$ intersect $AB$ at $X,Y$ respectively,prove that:

$\frac{1}{PY}+\frac{1}{PA}=\frac{1}{PX}+\frac{1}{PB}$

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pardesi

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pardesi

#4 h Oct 3, 2007, 1:49 AM • 2 Y

Y by Adventure10, Mango247

Thanks Babak. Do you know some more problems which can be proved similarly?

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BaBaK Ghalebi

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BaBaK Ghalebi

#5 h Oct 3, 2007, 6:41 AM • 9 Y

Y by Samurott, thunderz28, Adventure10, and 6 other users

pardesi wrote:

Thanks Babak. Do you know some more problems which can be proved similarly?

here you go,the first 5 problems:
problem 1
problem 2
problem 3
problem 4
problem 5

hope you enjoy them

This post has been edited 1 time. Last edited by BaBaK Ghalebi, Oct 3, 2007, 4:50 PM

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nsato

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nsato

#6 h Oct 3, 2007, 2:37 PM • 2 Y

Y by Adventure10, Mango247

BaBaK Ghalebi wrote:

problem 4

This problem does not use $Y$ . Is it stated correctly?

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Erken

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Erken

#7 h Oct 3, 2007, 3:11 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

BaBaK Ghalebi wrote:

Let $D,E,F$ be three points on the sides $BC,CA,AB$ of triangle $\triangle ABC$ such that $AD,BE,CF$ are concurrent at point $P$ ,let $EF$ intersect $AD$ at $K$ prove that $(A,K,P,D) = - 1$

Denote $EF\cap BC = K'$ ,since $(K',B,D,C) = - 1$ then the pencil $F(K',B,D,C)$ is harmonic $\Rightarrow(A,K,P,D) = - 1$
I think that it will be better if we will write $(A,B,D,C) = - 1$ ,than $(AC,BD) = - 1$ .

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Erken

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Erken

#8 h Oct 3, 2007, 3:37 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

BaBaK Ghalebi wrote:

Let $D,E,F$ be three points on the sides $BC,CA,AB$ of triangle $\triangle ABC$ such that $AD,BE,CF$ are concurrent at point $K$ ,let $FD$ intersect $BE$ at $X$ ,let $P$ be the midpoint of $AK$ and let $EP$ intersect $AB$ at $Y$ .prove that $XY\parallel AD$

I'll change the problem a little.Let $Y$ be the point on $AB$ such that $XY\|AD$ and $Z\in XY\cap AC$ .
Then we only need to prove that $Y$ is the midpoint of $XZ$ ,which is equivalent to(from Menelaus theorem):
$\frac{XB}{BE}=\frac{AZ}{AE}$ ,
since $XY\|AD$ then $\frac{AZ}{AE}=\frac{KX}{KE}$ ,but from the first problem we know that $(B,X,K,E)=-1\Rightarrow\frac{KX}{KE}=\frac{XB}{BE}$
Thus we are done.

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Erken

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Erken

#9 h Oct 3, 2007, 3:51 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

BaBaK Ghalebi wrote:

Let $D$ be a point on side $BC$ of triangle $\triangle ABC$ such that $AD\perp BC$ and let $P$ be an arbitary point on $AD$ ,let $BP,CP$ intersect $AC,AB$ respectively at $Y,X$ prove that6 $\angle XDA = \angle ADY$

Lemma
We will use one famous lemma:
Let $\angle ABC = 90^{\circ}$ and $X,Y\in AC$ ,then $(X,A,Y,C) = - 1\Longleftrightarrow \angle XBA = \angle ABY$ .
If someone want i could post the proof for this lemma,but it is too easy.
Solution(problem 3):
Let $XD\cap AC = Z$ then the division $(Z,A,Y,C)$ is harmonic,then using lemma we obtain that $\angle YDA = \angle ADZ = \angle ADX$ .Thus we are done.

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BaBaK Ghalebi

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BaBaK Ghalebi

#10 h Oct 3, 2007, 4:52 PM • 2 Y

Y by Adventure10 and 1 other user

nsato wrote:

BaBaK Ghalebi wrote:

problem 4

This problem does not use $Y$ . Is it stated correctly?

no actually...

its corrected now

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BaBaK Ghalebi

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BaBaK Ghalebi

#11 h Oct 3, 2007, 5:02 PM • 2 Y

Y by Adventure10 and 1 other user

Erken wrote:

BaBaK Ghalebi wrote:

Let $D,E,F$ be three points on the sides $BC,CA,AB$ of triangle $\triangle ABC$ such that $AD,BE,CF$ are concurrent at point $K$ ,let $FD$ intersect $BE$ at $X$ ,let $P$ be the midpoint of $AK$ and let $EP$ intersect $AB$ at $Y$ .prove that $XY\parallel AD$

I'll change the problem a little.Let $Y$ be the point on $AB$ such that $XY\|AD$ and $Z\in XY\cap AC$ .
Then we only need to prove that $Y$ is the midpoint of $XZ$ ,which is equivalent to(from Menelaus theorem):
$\frac {XB}{BE} = \frac {AZ}{AE}$ ,
since $XY\|AD$ then $\frac {AZ}{AE} = \frac {KX}{KE}$ ,but from the first problem we know that $(B,X,K,E) = - 1\Rightarrow\frac {KX}{KE} = \frac {XB}{BE}$
Thus we are done.

this one has an easier solution:
according to problem 1 we get that $(BK,XE) = - 1$ ,let $X'$ be the intersection of $XY,AD$ .now $YB,YK,YX,YE$ intersect line $AD$ at $A,K,X',P$ respectively,so $(AK,X'P) = - 1$ ,but we know that $P$ is the midpoint of $AK$ thus we must have $X'\to\infty$ i.e. $XY\parallel AD$ .
as wanted...

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BaBaK Ghalebi

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BaBaK Ghalebi

#12 h Oct 3, 2007, 5:18 PM • 1 Y

Y by Adventure10

Erken wrote:

I think that it will be better if we will write $(A,B,D,C) = - 1$ ,than $(AC,BD) = - 1$ .

why?
I always write it in that way...
it doesnt really matter,does it?

but actually in my notation from $(AC,BD)=k$ we mean that segment $AC$ has been devided internaly and externaly by $B,D$ .therefore we dont put "," between every two points...
anyway I dont think it makes any difference...

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Erken

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Erken

#13 h Oct 4, 2007, 10:29 AM • 1 Y

Y by Adventure10

BaBaK Ghalebi wrote:

Erken wrote:

I think that it will be better if we will write $(A,B,D,C) = - 1$ ,than $(AC,BD) = - 1$ .

why?
I always write it in that way...
it doesnt really matter,does it?

but actually in my notation from $(AC,BD) = k$ we mean that segment $AC$ has been devided internaly and externaly by $B,D$ .therefore we dont put "," between every two points...
anyway I dont think it makes any difference...

Yes,there is no diffrence.It is the first time,when i saw such notation,so excuse me,because as you said it desn't matter.
P.S Is this problem your owns?(Especcially 5th)
Thank you

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BaBaK Ghalebi

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BaBaK Ghalebi

#14 h Oct 4, 2007, 10:43 AM • 2 Y

Y by Adventure10, Mango247

Erken wrote:

Is this problem your owns?(Especcially 5th)
Thank you

no they're not,whats about the 5th one?

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Erken

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Erken

#15 h Oct 4, 2007, 11:11 AM • 2 Y

Y by Adventure10, Mango247

Oh,just the fifth problem is very beatiful,and i believe that there is a solution without using trig,i haven't still found it,i've found only one ugly solution with metrical relations.

I would like to ask you to not post your harmonic solution to this problem until tommorow.

OK?

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BaBaK Ghalebi

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BaBaK Ghalebi

#123 h Jan 2, 2008, 4:59 PM • 2 Y

Y by Adventure10, Mango247

unsolved problems wrote:

problem 7.

problem 8.

problem 11.

problem 12.

BaBaK

these are the unsolved problems of my last set...

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Erken

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Erken

#124 h Jan 3, 2008, 5:04 PM • 2 Y

Y by Adventure10, Mango247

BaBaK Ghalebi wrote:

let the incircle of triangle $\triangle ABC$ touch the sides $BC,CA,AB$ at $A',B',C'$ respectively,let $A''$ be a point on $BC$ such that $BA' = CA''$ now let $\ell_A$ be the line passing through $A'$ and parallel to $AA''$ ,define $\ell_B,\ell_C$ in the same way,prove that $\ell_A,\ell_B,\ell_C$ are concurrent.

I'm sorry,because i'll not use polarity and harmonic fours theory in my following solution,and also i'm afraid that polarity and "harmonic" solution is much shorter,but anyway here is my proof to this nice problem:
Solution:
Let $\omega(I,r)$ be the incircle of $\triangle ABC$ .Denote $B_1\in B'I\cap\omega$ (.Define $A_1,C_1$ in similar way.Let $l'_A,l'_B,l'_C$ be the tangent lines to $\omega$ at points $A_1,B_1,C_1$ respectively.Let $X,Y,Z$ be the points of their intesection,such that $X\in l'_C\cap l'_B,Y\in l'_C\cap l'_A,Z\in l'_A\cap l'_B$ .
It is well-known that $B,B_1,B''$ are collinear(similarly for other triplets of points) and it is easy to understand that $XI = AI,YI = BI,ZI = CI$ (from symmetry),
and it is obvious that $C'I = IC_1 = r$ ,hence $ZC'CC_1$ is parallelogramm $\Rightarrow$ $ZC'\|CC_1$ ,so $ZC'\in l_C$ .
Now,all we need to prove is the collinearity of $XA',YB',ZC'$ .But triangles $\triangle XYZ$ and $\triangle ABC$ are equal,and $XY\|AB,YZ\|BC,XZ\|AC$ ,hence
collinearity of $XA',YB',ZC'\Longleftrightarrow$ to the collinearity of $BB_1,CC_1,AA_1$ ,which is obviously true,and btw their intersection point is called the Nagel point.
Thus we are done

.

Remark:If we denote $T\in XA'\cap YB'\cap ZC'$ .From my solution it is easy to understand that, the point $I$ is the homothety center,which turns $\triangle XYZ$ to $\triangle ABC$ ,so the points $T,I,N$ are collinear,where $N$ is Nagel point,and $TI = IN$ .

Best regards,Erken.
P.S:To BaBaK Ghalebi:
Thank you for your nice problems.And please don't post solutions until next week,OK?
But i'm waiting for your proof of this problem,which uses polarity theory and harmonic fours.

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Erken

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Erken

#125 h Jan 8, 2008, 10:11 AM • 2 Y

Y by Adventure10, Mango247

Proof of Desarqeus Theorem with using :
The Desarques theorem:
Let $ABC$ and $A'B'C'$ be the 2 triangles such that $AA',BB',CC'$ are concurrent(at point $T$ ),
then points $X\in AB\cap A'B',Y\in AC\cap A'C',Z\in BC\cap B'C'$ are collinear.
Let A,B,C,D be the points on a line,which lies in that order,then let
$(AC,BD) = \frac {CA}{CB}\cdot\frac {DB}{DA}$ .
Proof:
Consider the line $l$ through the points $X$ and $Y$ .Denote:
$P\in AA'\cap l,Q\in BB'\cap l,R\in CC'cap l$ ,since
$PQ$ , $AB$ , $A'B'$ are concurrent then
$(TA,PA') = (TB,QB')$ .
Similarly,since $PR$ , $AC$ , $A'C'$ are concurrent then
$(TA,PA') = (TC,RC')$ .
Hence we conclude that
$(TB,QB') = (TC,RC')$ ,so
$BC,RQ,B'C'$ are concurrent,so we are done.

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darij grinberg

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darij grinberg

#126 h Jan 8, 2008, 11:00 AM • 2 Y

Y by Adventure10, Mango247

BaBaK Ghalebi wrote:

unsolved problems wrote:

problem 7.

problem 8.

problem 11.

problem 12.

BaBaK

these are the unsolved problems of my last set...

Are you sure you have proofs of all of these using harmonic conjugation? I am wondering what your proof for 12 looks like, since my approach is absolutely straightforward application of homotheties and reflections in points: A' is the reflection of A'' in the midpoint of BC, so if the parallel to AA'' through A'' and cyclically concur (what they surely do - at the Nagel point) and if the parallel to AA'' through the midpoint of BC and cyclically concur (what they also do, because the midpoints of BC, CA, AB form a triangle homothetic to ABC), then the parallel to AA'' through A' and cyclically must also concur. How do you use harmonic conjugation here? If you don't, please don't stuff these problems all into one thread.

Darij

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vittasko

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vittasko

#127 h Jan 8, 2008, 5:07 PM • 2 Y

Y by Adventure10, Mango247

darij grinberg wrote:

....Are you sure you have proofs of all of these using harmonic conjugation? I am wondering what your proof for 12 looks like, since my approach is absolutely straightforward application of homotheties and reflections in points: A' is the reflection of A'' in the midpoint of BC, so if the parallel to AA'' through A'' and cyclically concur (what they surely do - at the Nagel point) and if the parallel to AA'' through the midpoint of BC and cyclically concur (what they also do, because the midpoints of BC, CA, AB form a triangle homothetic to ABC), then the parallel to AA'' through A' and cyclically must also concur....

Alternatively, I think it is well known (or easy to prove) that the lines through the midpoints of the side-segments of the given triangle $\bigtriangleup ABC,$ and parallel to $AA'',$ $BB'',$ $CC'',$ pass through its incenter $I$ and so, we can easily conclude, by Thales theoerem, the concurrency of the lines through $A',$ $B',$ $C'$ and parallel to the line segments $AA'',$ $BB'',$ $CC''$ respectively, at one point, as the reflexion of the Nagel point $N\equiv AA''\cap BB''\cap CC'',$ with respect to the incenter $I$ of $\bigtriangleup ABC.$

Kostas Vittas.

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pardesi

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pardesi

#128 h Jan 17, 2008, 3:34 PM • 1 Y

Y by Adventure10

BaBaK Ghalebi wrote:

Erken wrote:

BaBaK Ghalebi wrote:

Let $w$ be a semicircle with diameter $AB$ and let $C$ be a point on $w$ .assume that the tangents from points $A,C$ to $w$ intersect at $D$ .let $H$ be a point on $AB$ such that $CH\perp AB$ and let $CH$ intersect $BD$ at $K$ .prove that $HK = KC$

Denote $F\in w\cap BD$ , $G\in AD\cap BC$ .Then $AFCB$ is harmonic quadrilateral,hence $B(BCFA)\cap AD$ makes a harmonic four,hence $AD = DG$ .Now using that $AB$ is diameter,we obtain that $AG\|CH$ ,hence $CK = KH$ .Thus we are done.

my solution to this problem is somehow the same,anyway here it goes:
first of all we have the following theorem:
for point $P$ and circle $w$ ,let $M,N$ be points on $w$ such that $PM,PN$ are tangent to $w$ .now let the chord $AB$ of $w$ pass through $P$ ,then $(AB,MN) = - 1$
now in our problem let $BD$ intersect $w$ at $P$ ,then according to the lemma above we get that $(AC,PB) = - 1$ now denote $\ell$ as the tangent to $w$ at $B$ .
now we know that $BA,BC,BP$ intersect line $CH$ at $H,C,K$ respectively and also $B$ lies on $w$ so $(HC,KB') = - 1$ where $B' = \ell\cap CH$ but in the other hand we know that $\ell\perp AB$ and $CH\perp AB$ thus $\ell\parallel CH$ i.e. $B'\to\infty$ so:
$(HC,K\infty) = - 1$ or in other words $K$ is the midpoint of $HC$ i.e. $HK = KC$ .

BaBak i don't understand the part where B is projected about itself...is that possible and why should the line koining B to B we the line $l$ only it could be any other line through B

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BaBaK Ghalebi

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BaBaK Ghalebi

#129 h Jan 18, 2008, 10:21 AM • 1 Y

Y by Adventure10

pardesi,let $B$ be a fixed point on circle $w$ and $A$ be a variant point on $w$ ,now let $\ell$ be the line passing through $A,B$ so $\ell$ variates as $A$ variates...
but note that $\ell$ always intersects $w$ at points $A,B$
now in the case $A\to B$ line $\ell$ intersects circle $w$ at $A,B$ where $A\equiv B$ so $\ell$ intersects $w$ at a single point $B$ i.e. $\ell$ is tangent to $w$ at point $B$ ...
so the line passing through $B$ and connecting it to $B$ is the line tangent to $w$ at $B$ i.e. the line $BB$ is tangent to $w$ at point $B$ ...

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pardesi

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pardesi

#130 h Jan 18, 2008, 1:45 PM • 2 Y

Y by Adventure10, Mango247

thanks BaBaK...

also just another doubt when $A,B,C,D$ are concyclic and we write $(AB,CD)=-1$ ..then do we mean if they are projected about any point $P$ onto a line then their projections form harmonic

:

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BaBaK Ghalebi

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BaBaK Ghalebi

#131 h Jan 18, 2008, 5:09 PM • 3 Y

Y by Adventure10, Mango247, and 1 other user

let $P,A,B,C,D$ be points on circle $w$ ,let $PA,PB,PC,PD$ intersect line $\ell$ at points $A',B',C',D'$ now assume that $(AB,CD) = k$ i.e. $\frac {AC}{BC} = k\cdot\frac {AD}{BD}$ ...
now note that:

$\frac {AC}{BC} = k\cdot\frac {AD}{BD}\iff \frac {\frac {AC}{2R}}{\frac {BC}{2R}} = k\cdot\frac {\frac {AD}{2R}}{\frac {BD}{2R}}$ (*)

now note that we have $\frac {AC}{2R} = \sin \angle APC$ and so on,so the equation (*) is equivalent to:

$(*)\iff\frac {\sin \angle APC}{\sin \angle CPB} = k\cdot\frac {\sin \angle APD}{\sin \angle DPB}$

now according to the $\sin$ law in triangles $\triangle PA'C',\triangle PC'B'$ we get that:

$\frac {\sin \angle APC}{\sin \angle CPB} = \frac {\sin \angle A'PC'}{\sin \angle C'PB'} = \frac {A'C'}{B'C'}$

similary in triangles $\triangle PA'D',\triangle PD'B'$ we have:

$\frac {\sin \angle APD}{\sin \angle DPB} = \frac {\sin \angle A'PD'}{\sin \angle D'PB'} = \frac {A'D'}{D'B'}$

hence by (*) we get that:

$\frac {A'C'}{B'C'} = k\cdot\frac {A'D'}{B'D'}\Rightarrow (A'B',C'D') = k$

$\Rightarrow (AB,CD) = k\iff (A'B',C'D') = k$

as wanted...

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pardesi

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pardesi

#132 h Jan 18, 2008, 5:15 PM • 2 Y

Y by Adventure10, Mango247

thank u BaBaK

i had a seriuos misconception.

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Erken

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Erken

#133 h Mar 2, 2008, 3:06 PM • 2 Y

Y by Adventure10, Mango247

Let $\omega$ be a circle and $AB,CD$ be two parallel chords of $\omega$ .Let $M$ be the midpoint of $AB$ ,and $E$ be the second intersection of $CM$ with $\omega$ .Prove that $\angle AKE=\angle EKB$ ,where $K$ is the midpoint of $DE$ .

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BaBaK Ghalebi

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BaBaK Ghalebi

#134 h Mar 3, 2008, 7:35 PM • 2 Y

Y by Adventure10, Mango247

Erken wrote:

Let $\omega$ be a circle and $AB,CD$ be two parallel chords of $\omega$ .Let $M$ be the midpoint of $AB$ ,and $E$ be the second intersection of $CM$ with $\omega$ .Prove that $\angle AKE = \angle EKB$ ,where $K$ is the midpoint of $DE$ .

nice problem.

let $S$ be the intersection point of $DE,AB$ ,now note that $M$ is the midpoint of $AB$ so $(AB,M\infty)=-1$ and the lines $CA,CB,CM$ intersects the circle at $A,B,E$ respectively and also the parallel line with $AB$ ,through $C$ ,intersects the circle at $D$ so:

$(AB,M\infty)=-1\Rightarrow (AB,ED)=-1$

now let the tangent line to the circle at point $D$ intersect $AB$ at point $T$ then we have:

$(AB,ED)=-1\Rightarrow (AB,ST)=-1$

so in order to prove $\angle AKE=\angle EKB$ we have to prove that $\angle TKS=90^\circ$ :

as we said before $(AB,ED)=-1$ so $AB$ must pass through the intersection of the tangent lines at $D,E$ .so the tangent to $w$ at $E$ passes through $T$ i.e. $TE$ is tangent to $w$ .now note that $TE,TD$ are two tangents from point $T$ to circle $w$ so $TE=TD$ so $\triangle ETD$ is isosceles,but on the other hand $K$ is the midpoint of $ED$ thus:

$TK\mid ED\Rightarrow \angle TKE=90^\circ$ .

QED

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Erken

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Erken

#135 h Mar 4, 2008, 1:42 AM • 2 Y

Y by Adventure10, Mango247

Great,but another easier solution,could be obtained,if we use the fact,that dioganal is the symmedian in harmonic quadrilateral.

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Enjar0

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Enjar0

#136 h May 22, 2008, 1:33 PM • 3 Y

Y by Adventure10, Adventure10, Mango247

Erken wrote:

BaBaK Ghalebi wrote:

Let $D$ be a point on side $BC$ of triangle $\triangle ABC$ such that $AD\perp BC$ and let $P$ be an arbitary point on $AD$ ,let $BP,CP$ intersect $AC,AB$ respectively at $Y,X$ prove that6 $\angle XDA = \angle ADY$

Lemma
We will use one famous lemma:
Let $\angle ABC = 90^{\circ}$ and $X,Y\in AC$ ,then $(X,A,Y,C) = - 1\Longleftrightarrow \angle XBA = \angle ABY$ .
If someone want i could post the proof for this lemma,but it is too easy.
Solution(problem 3):
Let $XD\cap AC = Z$ then the division $(Z,A,Y,C)$ is harmonic,then using lemma we obtain that $\angle YDA = \angle ADZ = \angle ADX$ .Thus we are done.

I want Erken pls

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math1990

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math1990

#137 h Jul 10, 2008, 6:28 PM • 2 Y

Y by Adventure10, Mango247

Hi everyone ! I can't understand this "Harmonic Division" what this mean? I have seen about harmonic division at "Mathematical reflections" but Can't understand.

:

what this mean (CA/CB=-DA/DB)
and what this mean (AB,CD)=-1 and (AB,CD)=k ?

please write full about it !

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