cyclic quads to parallelogram

[Tahmid Hasan]

Let ABCD be a cyclic quadrilateral whose opposite sides are not parallel, X the intersection of AB and CD, and Y the intersection of AD and BC. Let the angle bisector of ∠AXD intersect AD, BC at E, F respectively and let the angle bisector of ∠AYB intersect AB, CD at G, H respectively. Prove that EGFH is a parallelogram.

[Zzzz]

Please... someone try to to solve the problem (I am asking junior and secondary level students).

There is a hint about the problem:

Draw the figure using ruler and compass. Surely you will find something interesting. Not only for this problem, after drawing the figure (properly) of almost any geometry problem, you will find something interesting

[photon]

mine is different with points according to the problem.Let the angle bisector of ∠AXD intersect AD, BC at S, Q respectively and let the angle bisector of ∠AYB intersect AB, CD at P, R respectively.
$\angle A=180^O-\angle C=\angle QCX$
$\angle AXS=\angle CXQ=\frac{1}{2}\angle X $
THUS,$\bigtriangleup AXS\sim \bigtriangleup CQX$
$\angle XSA=\angle CQX$
$so,\angle QSD=\angle CQS$
in, $\bigtriangleup YQS,\frac{sin\angle QSY}{sin\angle SQY}=\frac{QY}{SY}$
so,$YQ=YS$
THEN BY ANGLE BI-SECTING THEOREM,$OQ=OS$
again,we can get $XP=XR$ according to above way and will get $OP=OR$
in $PQRS$, $OQ=OS$ and $OP=OR$,we can easily show two oppsite sides parallel and equal by congruent triangles.

[Phlembac Adib Hasan]

Proof:

Join $A,C$ and $B,D$.
From power of point, \[ \frac {CX} {BX}=\frac {AX} {DX} \]
\[\Rightarrow \frac {CF} {FB}= \frac {AE} {ED} \]
Similarly, \[\frac {AG} {GB}=\frac {CH} {HD} \]
So, we can say that, \[\frac {CH} {CF}=\frac {AG} {GB}. \frac {DE} {AE} .\frac {HD} {FB} \]
\[\Rightarrow \frac {CH} {CF}=\frac {AY} {YB} .\frac {DX} {AX} .\frac {HD} {FB} \]
Notice that $ \triangle DXB $ and $ \triangle AXC $ Similar.So,$\frac{DX}{AX}=\frac {BD} {AC}$.Again from similarity of $ \triangle DBY $ and $ \triangle CAY $, $\frac {AY} {BY}=\frac {AC} {BD}$.
So, \[ \frac {CH} {CF}= \frac {AG} {GB}. \frac {DE} {AE} .\frac {HD} {FB}= \frac {AC} {BD}. \frac {BD} {AC} .\frac {HD} {FB}\]
\[ \Rightarrow \frac {CH} {CF}=\frac {HD} {FB}=\frac{CH+HD} {CF+FB}=\frac {CD} {BC} \]
So, $ \triangle CHF $ and $ \triangle CDB $ Similar.It implies $FH ||BD$.
Similar argument shows that $EG||BD$ means $FH||EG$.Similarly, $EH||GF$.So $EHFG$ is a parallelogram.

[sourav das]

Actually the fact that almost kill the problem is to prove $EFGH$ is a rohmbos

[Phlembac Adib Hasan]

My previous solution was not complete. Additional part:

From similarity of triangles $ \triangle DXB $ and $ \triangle AXC, $

$ \frac {AC} {BD}= \frac {AX} {DX} = \frac {AE} {DE} $ [angle bisector theorem]

$ \Rightarrow AC.DE=BD.AE $

$ \Rightarrow \frac {AC.DE} {AD}= \frac {BD.AE} {AD} $

$ \Rightarrow EH=EG $.And we are done.