ISL 2006

[shehab ahmed]

Let ABCDE be a convex pentagon such that
$\angle BAC= \angle CAD= \angle DAE$ and
$ \angle ABC= \angle ACD= \angle ADE$
If $BD$ and $CE$ intersect at $P$,then prove that $AP$ bisects $CD$.

[SANZEED]

It is easy to notice that $BC$ is a tangent to the circumcircle of \[\triangle ACD\].So is $ED$.Again note that $CD$ is a tangent the circumcircle of \[\triangle ADE\] and to that of \[\triangle ABC\].Since \[\angle BAC=\angle DAE\] and \[\angle ABC=\angle ADE\],we can say that \[\triangle ABC\sim \triangle ADE\].Again,\[\angle EAC=\angle EAD+\angle DAC=\angle DAC+\angle CAB=\angle DAB\].Both of this together implies that a spiral similarity takes \[\triangle AEC\] to \[\triangle ADB\] and both $A,P$ lies on the circumcircles of \[\triangle ADE,\triangle ABC\].Since $AP$ is the radical axis of these circles,it bisects their common tangent $CD$.

[Phlembac Adib Hasan]

My Proof:

Firstly I'd like to define some things:
$AP\cap CD=M,\angle BAC=x,\angle ABC=y,\angle ACB=z,AC\cap BD=X,AD\cap CE=Y.$
Notice that a spiral similarity takes $\triangle ABC$ to $\triangle ADE$.Now prove $\triangle ABD\sim \triangle ACE$ i.e $ABCP$ and $AEDP$ concyclic.So $\angle BPC=\angle XAY=x$Hence $AXPY$ also concyclic.Again $\angle AXY=\angle APY=\angle APE=\angle ADE=y$.So it follows $\triangle AXY\sim \triangle ACD$.Hence $\frac {AX}{AY}=\frac {AC-AX}{AD-AY}=\frac {CX}{DY}$.Now last part can be handled easily by Ceva\[\frac {CM}{MD}=\frac {CX}{AX}\frac {AY}{DY}=1\]Therefore $M$ is the mid-point of $CD$.

(I am really very grateful to Tahmid Vaia for the last part.Go to this link for details:viewtopic.php?f=25&t=2075[Read his sketch of solution])

[zadid xcalibured]

One can use similar quadrilaterals and ceva's theorem.