Cyclic quadrilater

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Source: Romanian DMO 9th grade P4

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#1 h Mar 1, 2008, 4:15 PM • 2 Y

Y by Adventure10, Mango247

Let $ABCD$ be a cyclic quadrilater. Denote $P=AD\cap BC$ and $Q=AB \cap CD$ . Let $E$ be the fourth vertex of the parallelogram $ABCE$ and $F=CE\cap PQ$ . Prove that $D,E,F$ and $Q$ lie on the same circle.

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Ashegh

#2 h Mar 2, 2008, 8:24 AM • 2 Y

Y by Adventure10, Mango247

i want to show that: $CD.CQ=CE.CF$ (1)

then i will find these length,according to their relations to each other:

according to a lemma in inversion,if i choose $P$ ,the center,and power of $P$ to circle of the quadrilateral,as the radius $K$ ,

i can find the length of $CD$ this way:

$CD=\frac{k.AB}{PA.PB}=\frac{PB.PC.AB}{PA.PB}=\frac{PC.PB}{PA}$

$ABCE$ is paralelogram,then: $CE=AB$

we know that: $QC.QD=QB.QA$ ,then: $QC=\frac{QB.QA}{QD}$

triangles $PBQ,PCF$ are similar to each other,then: $\frac{CF}{BQ}=\frac{PC}{PB}$ ,

then we conclude: $CF=\frac{PC.BQ}{PB}$

now we find the length,lets use them in (1)equality,and check if it is right or not.

$\frac{PC.AB}{PA}.\frac{QB.QA}{QD}=AB.\frac{PC.BQ}{PB}$

it concludes that,we should have: $\frac{QA}{PA.QD}=\frac{1}{PB}$

$\frac{PA}{PB}=\frac{PC}{PD}$

now i should chek that if it is right or not.

ok,it is clear that $\angle PCQ=\angle QAD$

$\sin{PCQ}=\sin(180-QAD)$

suppose $PH,QH'$ are the altitudes of triangle $PDQ$ .

we can find the area of $PDQ$ ,in two ways:

$S=PH.QD=QH'.PD$

but: $PH=PC.\sin{PCQ}$ and $QH'=QA.\sin(180-QAD)$

and we conclude that: $PC.QD=QA.PD$ and it is proved that: $\frac{PA}{PB}=\frac{QA}{QD}$

and every thing is ok,but we should go back these steps from end to first,

and then the proof will be complete

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#3 h Mar 2, 2008, 11:07 AM • 2 Y

Y by Adventure10, Mango247

Let $K,L,M$ be midpoint of $BD,CA,PQ$ . By Gauss line theorem(See http://www.mathlinks.ro/Forum/viewtopic.php?t=79751) , $K,L,M$ are collinear. Then reflection of $B$ in the point $K,L,M$ are also collinear, i. e, $D,E,X$ are collinear where $X$ is reflection of $B$ in the point $M$ . Since $XPBQ$ is parallelogram, $\angle QXP = \angle PBQ = \angle CBA = \angle CDA = \angle QDP$ . Hence $X,P,D,Q$ are cyclc. From $XPBQ$ and $ECBA$ are parallelogram, $XP\parallel EC$ . So $\angle DEF = \angle DXP = \angle DQP = \angle DQF$ . So $D,E,F,Q$ are cyclic.

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#4 h Mar 2, 2008, 1:57 PM • 2 Y

Y by Adventure10 and 1 other user

We draw the line through the point $Q$ and parallel to $BC,$ which intersects the line segment $EF$ at one point so be it $F'.$

It is enough to prove that $CD\cdot CQ = CE\cdot CF$ $\Longrightarrow$ $CD\cdot CQ = AB\cdot CF$ $,(1)$

From $E'F\parallel AQ,$ where $E'\equiv AD\cap EF,$ we have by Thales theorem, $\frac {AB}{BQ} = \frac {CE'}{CF}$ $\Longrightarrow$ $AB\cdot CF = BQ\cdot CE'$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $CD\cdot CQ = BQ\cdot CE'$ $,(3)$ and so, it is enough to prove that $CD\cdot CQ = CF'\cdot CE'$ $,(4)$

But, the relation $(4)$ is true, from the cyclic quadrilateral $DE'QF',$ because of

$\angle DE'F' = \angle DAB = \angle BCQ = \angle DQF'$ $\Longrightarrow$ $\angle DE'F' = \angle DQF'$ $,(5)$

Hence, we conclude that $CD\cdot CQ = CE\cdot CF$ and the proof is completed.

Kostas Vittas.

Attachments:

t=191681.pdf (6kb)

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#5 h Mar 6, 2008, 7:37 PM • 2 Y

Y by Adventure10, Mango247

vittasko wrote:

We draw the line through the point $Q$ and parallel to $BC,$ which intersects the line segment $EF$ at one point so be it $F'.$

It is enough to prove that $CD\cdot CQ = CE\cdot CF$ $\Longrightarrow$ $CD\cdot CQ = AB\cdot CF$ $,(1)$

From $E'F\parallel AQ,$ where $E'\equiv AD\cap EF,$ we have by Thales theorem, $\frac {AB}{BQ} = \frac {CE'}{CF}$ $\Longrightarrow$ $AB\cdot CF = BQ\cdot CE'$ $,(2)$

From $(1),$ $(2)$ $\Longrightarrow$ $CD\cdot CQ = BQ\cdot CE'$ $,(3)$ and so, it is enough to prove that $CD\cdot CQ = CF'\cdot CE'$ $,(4)$

But, the relation $(4)$ is true, from the cyclic quadrilateral $DE'QF',$ because of

$\angle DE'F' = \angle DAB = \angle BCQ = \angle DQF'$ $\Longrightarrow$ $\angle DE'F' = \angle DQF'$ $,(5)$

Hence, we conclude that $CD\cdot CQ = CE\cdot CF$ and the proof is completed.

Kostas Vittas.

Dear Kostas,

nice solution ! In the official solution ( very comprehensive )I read the relation :

$\frac { CP}{DP} = \frac {CQ}{BQ}$

How can we deduce this relation without trigononetry ?

babis

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#6 h Mar 6, 2008, 8:43 PM • 2 Y

Y by Adventure10, Mango247

stergiu wrote:

....
Dear Kostas, In the official solution ( very comprehensive )I read the relation :

$\frac { CP}{DP} = \frac {CQ}{BQ}$

How can we deduce this relation without trigononetry ?

babis

Because of the triangles $($ schema t=191681 $)$ $\bigtriangleup PDC$ and $\bigtriangleup QBC,$ have $\angle PCD = \angle QCB$ and $\angle PDC + \angle QBC = 180^{o},$

we conclude that $\frac {CP}{CQ} = \frac {DP}{BQ}$

$($ The ratio of their side-segments opposite to their equal angles, is congruent to the ratio of their side-segments opposite to their angles with sum $180^{o}$ $).$

But, how is the official solution you mentioned? Have you any reference about it?

Best regards, Kostas vittas.

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#7 h Mar 7, 2008, 8:59 PM • 2 Y

Y by Adventure10, Mango247

vittasko wrote:

stergiu wrote:

....
Dear Kostas, In the official solution ( very comprehensive )I read the relation :

$\frac { CP}{DP} = \frac {CQ}{BQ}$

How can we deduce this relation without trigononetry ?

babis

Because of the triangles $($ schema t=191681 $)$ $\bigtriangleup PDC$ and $\bigtriangleup QBC,$ have $\angle PCD = \angle QCB$ and $\angle PDC + \angle QBC = 180^{o},$

we conclude that $\frac {CP}{CQ} = \frac {DP}{BQ}$

$($ The ratio of their side-segments opposite to their equal angles, is congruent to the ratio of their side-segments opposite to their angles with sum $180^{o}$ $).$

But, how is the official solution you mentioned? Have you any reference about it?

Best regards, Kostas vittas.

I had found a link in romanian languange, but I have lost it. I will have the official solution next year , when I will buy the book with all math olympiads for year 2008.
Babis

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#8 h Mar 8, 2008, 5:47 PM • 2 Y

Y by Adventure10, Mango247

The official solution is based on some raport equality and one Menelaos.

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#9 h Mar 9, 2008, 10:28 AM • 1 Y

Y by Adventure10

Let $AE$ meet $CD$ at $K$
The line parallel with $QC$ through $P$ meet $BQ$ at $M$ $(1)$
We have $\angle KCE=\angle AQD=\angle PMA$ (because $CE//BQ$ and (1))
and $\angle PAM=\angle DCB=\angle EKC$ ( because $A,B,C,D$ are cyclic and $AE//BC$ )
Thus triangles $PMA$ and $ECK$ are similar $(2)$
We have $CK/CD=PA/PD=MA/MQ$ $(3)$
From (2) and (3) $CE/MP=CK/MA=CD/MQ$
Thus triangles $PMQ$ and $ECD$ are similar
So $\angle EDC=\angle PQM=\angle QFC$
Hence $Q,D,E,F$ are cyclic $q.e.d$

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