we have (from similarity of ABC, ACD,ADE): $\frac{AB}{BC}=\frac{AC}{DC}=\frac{AD}{ED}$ and $\frac{BC}{AC}=\frac{CD}{AD}=\frac{ED}{AE}$ from which we obtain: $AB \cdot AD =AC^{2}$ and $AC \cdot AE = AD^{2}$ and hence $AB \cdot AE = AC \cdot AD$ or otherwise $\frac{AE}{AC}=\frac{AD}{AB}$ hence triangles $ABD$ and $AEC$ are similar. Now it's just angle chasing. Use this similarity to have some equal angles and see that $ABCP$ and $AEDP$ are cyclic and tangent to line $CD$ respectively at $C$ and $D$. Hence if $M$ is intersection of AP and CD then $MD^{2}=MP \cdot MA = MC^{2}$ so $M$ is midpoint of $CD$ as wanted. QED

if the quadrilaterals ABCP and AEDP are cyclic then ADB and AEC are isoceles. am i missing something???

Hasan, I don't think that the triangles you mention are isosceles. Please have a look at my picture. Megus, I really like your proof. It is very nice! I found two proofs of this nice and easy problem. Look at the picture for names of points. Proof 1: A very short, straightforward one: Clearly, there exists a spiral similarity with center $A$, angle $BAC$ and factor $\frac{AC}{AB}$ that maps $ABCD$ onto $ACDE$. Because $F$ and $G$ are the intersection points of the diagonals of two quadrilaterals that are similar, we know that \[\frac{\overline{FA}}{\overline{FC}}=\frac{\overline{GA}}{\overline{GC}}\] So if we call $\left\{S\right\}=AP \cap CD$, we know thanks to Ceva's theorem: \[-1=\frac{\overline{FA}}{\overline{FC}}.\frac{\overline{SC}}{\overline{SD}}.\frac{\overline{GD}}{\overline{GA}}=\frac{\overline{SC}}{\overline{SD}}\] So clearly $S$ is the midpoint of $CD$ Proof 2: My second proof, which I think is more beautiful: Clearly, $\Delta ABC, \Delta ACD, \Delta ADE$ are similar. Therefor, we can use the alternating segment theorem to prove that both $BC$ and $DE$ are tangent to the circumcircle of $\Delta ACD$ in points $C$ and $D$ respectively. If we call $Q$ the intersection of $BC$ and $DE$, we know that $AQ$ is a symmedian of $\Delta ACD$. Now, because $\angle ACB = \angle AED$, we know that $ACQE$ is cyclic and $\angle CAQ = \angle CEQ$ (1) Because $\angle ADB = \angle AEC$, we know that $AEDP$ is cyclic and $\angle PAD = \angle PED$ (2) Combining (1) and (2), we get that $AP$ is the isogonal conjugate of $AQ$, and therefor, it is the median. The conclusion follows. I hope you liked it Image not found

Thank you dear jan, for the enjoyment from the beautiful ''proof 2'', of the mattilgale's problem. Best regards, Kostas vittas.

Megus wrote: we have (from similarity of ABC, ACD,ADE): $\frac{AB}{BC}=\frac{AC}{DC}=\frac{AD}{ED}$ and $\frac{BC}{AC}=\frac{CD}{AD}=\frac{ED}{AE}$ from which we obtain: $AB \cdot AD =AC^{2}$ and $AC \cdot AE = AD^{2}$ and hence $AB \cdot AE = AC \cdot AD$ or otherwise $\frac{AE}{AC}=\frac{AD}{AB}$ hence triangles $ABD$ and $AEC$ are similar. Now it's just angle chasing. Use this similarity to have some equal angles and see that $ABCP$ and $AEDP$ are cyclic and tangent to line $CD$ respectively at $C$ and $D$. Hence if $M$ is intersection of AP and CD then $MD^{2}=MP \cdot MA = MC^{2}$ so $M$ is midpoint of $CD$ as wanted. QED As a remark to the Megus’s nice solution, we can say that the quadrilaterals $ABCD,$ $ACDE$ are similar, because of their angles, are equal one per one and so, we conclude that their homologous segment lines, form also equal angles. Hence, as another approach, we have $\angle ADB = \angle AEC$ $,(1)$ and $\angle BDC = \angle CED$ $,(2)$ From $(1)$ we conclude that the quadrilateral $APDE$ is cyclic and so, $\angle DPE = \angle DAE = \angle DAC$ $,(3)$ From $(3)$ we have that the quadrilateral $AFPG,$ is also cyclic. Now, it is easy to show that $\angle GFP = \angle PAD = \angle CED$ $,(4)$ From $(2),$ $(4)$ $\Longrightarrow$ $\angle GFP = \angle BDC$ $\Longrightarrow$ $FG\parallel CD$ $,(5)$ From trapezium $CDGF,$ we conclude that the segment line $AP$ bisects the segment $CD$ and the proof is completed. Kostas Vittas.

Dear Megus , can you please post your solution with a little more detail . especially the angle chasing part, ive tried to finish it but cant find how. thank you very much

vittasko wrote: As a remark to the Megus’s nice solution, we can say that the quadrilaterals $ABCD,$ $ACDE$ are similar, because of their angles, are equal one per one and so, we conclude that their homologous segment lines, form also equal angles. Also we can say that their homologous points, divide the correspondent segments in to the same ratio. That is from the similarity of $ABCD,$ $ACDE,$ as a direct conclusion we have that $\frac{FA}{FC}= \frac{GA}{GD}$ and so, $FG\parallel CD.$ Kostas Vittas.

Well, with simple angle chasing one can see that $ABCP$ and $APDE$ are cyclic and their radical axis is $AP$. It is also easy to see that $CD$ is a common tangent of circumcircles $\triangle ABC$ and $\triangle DEA$. Therefore $AP$ passes through a point M in CD such that $MC^{2}=MD^{2}$.

As Darij Grinberg said at http://www.mathlinks.ro/Forum/viewtopic.php?t=154396, we can have a solution as a neat application of Jacobi theorem. Through vertices $C,$ $D$ of the given pentagon $ABCDE,$ we draw two lines parallel to $AD,$ $AC$ respectively and we denote as $F,$ their intersection point. So, it is easy to show that the segment lines $DE,$ $DF,$ are isogonal conjugates with respect to the angle $\angle ADC$ and similarly, the segment lines $CB,$ $CF,$ are isogonal conjugates wrt $\angle ACD.$ Hence, because of $\angle DAE = \angle CAB,$ based on the Jacobi theorem, we conclude that $P\equiv AF\cap CE\cap DB.$ Because of now the parallelogram $ADFC,$ we have that the segment line $AP,$ passes through the midpoint of the side segment $CD$ and the proof is completed. Many thanks to Darij, Kostas Vittas.

I hope it is a nice solution, too Let's call the interception of the diagonals F and G and M the middle of CD Since ABC is similiar to ACD and ADE it is obvious that ABCD is similiar to ACDE Therefore AF/FC = AG/GD Ceva: AF/FC*CM/MD*DG/GA=1, which is obviously true

Call the intersection of (ABC) and (ADE) is X. easy to prove that X = P. We can prove that CD is common targent of (ABC) and (ADE). Hence, AP is axial radical of (ABC) and (ADE) so AP passes through the midpoint of CD

Since triangle $ABC$ is similar to triangle $ADE$, $A$ is the spiral centre which takes $B$ to $C$ and $D$ to $E$, hence by a well known lemma, the point of intersection of $CE$ and $BD$ is concyclic with $ABC$ and $ADE$. Further, by easy angle chasing and the alternate segment theorem, $CD$ is tangent to both of these circles, hence since the power of the point of intersection of $AP$ and $CD$, $M$, is equal to $MP.MA$ for both circles, it is immediate that $MC$=$MD$.

Let $R=AC \cap BD, Q=EC\cap AD, K=AE\cap CD, L=AD\cap BC$ and $M=AP \cap CD$ Ceva's theorem gives \[ \frac{MD}{MC}=\frac{QD}{QA}\frac{RA}{RC}\] Menelaus' theorem in $ACL$ with line $D,R,B$ gives \[\frac{RA}{RC}=\frac{BC.DL}{BL.AD}\] Menelaus' theorem in $ADK$ with line $E,Q,C$ gives \[\frac{QD}{QA}=\frac{CK.AE}{CD.KE}\] $DLC$ and $EKD$ are similar so \[\frac{DL}{KE}=\frac{CD}{DE}\] $ABC$ and $ADE$ are similar so \[\frac{BC}{DE}=\frac{AC}{AE}\] $ABL$ and $ACK$ are similar, $AC$ and $AD$ are respectively their angles bisector from $\hat{A}$ so \[\frac{AC}{BL}=\frac{AD}{CK}\] so \[\frac{MD}{MC}=\frac{RA}{RC}\frac{QD}{QA}=\frac{BC.DL}{BL.AD}\frac{CK.AE}{CD.KE}=...=\frac{AC.CK}{BL.AD}=1\] then $MC=MD$ as required

Am I the only one who feels that ISL 2006 is easier than other years Denote $AC \cap BD = X$ and $AD \cap CE=Y$ Thus we are required to prove that $\frac{AX}{CX}=\frac{AY}{DY}$ Then the result will follow immediately from the Ceva's Theorem. Let $\angle BAC=a$, $\angle ABC=b$ and $\angle ACB=c$ Then $\frac{AX}{CX}=\frac{[ABD]}{[BCD]}=\frac{AD \cdot AB \cdot \sin (2a)}{CD \cdot CB \cdot \sin(b+c)}$ Similarly, $\frac{AY}{CY}=\frac{AC \cdot AE \cdot \sin (2a)}{CD \cdot DE \cdot \sin(b+c)}$ Thus we need to prove $\frac{AD \cdot AB}{CB}=\frac{AC \cdot AE}{DE}$ This can be easily proved by using Sine Formula repeatedly

Though my proof is very similar to Megus' proof (actually the same) but still I'll post it. Denote by $S(O,\theta,k)$ denote the spiral symmetry with centre $O$, angle of rotation $\theta$ and ratio $k$ Lemma If a spiral symmetry maps $S \to R$ and $Q \to P$ and $P$ denote the intersection of $RS$ and $PQ$ $\implies$ the centre of spiral symmetry lies on the second intersection of the circumcircles of $BCP$ and $EDP$ In the given question, Let $\angle BAC = \theta$ Now it is easy to see that $S(A,2 \theta, \frac{AB}{AD})$ takes $B \to D$ and $C \to E$ Thus by our above lemma, $APCB, APDE$ are concyclic. Also it is easy to see that $CD$ is the common tangent of the circumcircles of the above two quadrilaterals. Let the intersection $AP$ and $CD$ be $X$ Then $XC^2=XP \cdot XA = XD^2$ $\implies XC=XD$ Thus the result

This is not at all a problem suitable for the shortlist.Ideal geometry problems should be hard to bash,but can be cut through by deep insights to the problem.This problem can be solved by judt using sine rule,as Utkarsh pointed out, in 4-5 triangles.I did this problem but will not post it here since it has the same approach as Utkarsh.

it could be solved easier after you prove that $ABCP$ and $AEDP$ are concyclic. From similarity of triangles $ABC$, $ACD$ and $ADE$ you get \[ \angle ABC = \angle ACD\]and \[ \angle ADC = \angle AED\]and then $CD$ is common tangent of this circles and since $AP$ is radical axis, $AP$ bisects $CD$.

I just wanted to post a different solution using isogonal lines lemma. It states that for a triangle ABC with pairwise isogonal lines AD-AG and AE-AF, then DE with FG and DF with EG intersecting at H and I yields that AH and AI are isogonal lines. Sketch. For J,K the intersections of AH with DF and EG, x=<PAS=<QAT, y=<SAX and z=<TAY, we have that sinKAP/sinKAY : sinTAP/sinTAY=(P,Y;K,T)=(S,Y;L,Q)=sinLAS/sinLAY : sinQAS/QAY after a flurry of simplifications reduces to sin(x+y)sinz=sin(x+z)siny, which reduces to tany=tanz or y=z. $\square$ Now, in the problem, AC-AD and AB-AE are isogonal line pairs, so for BD intersecting EG at F and DE intersecting BC at G, we get that AF-AG is an isogonal line pair as well (all wrt <CAD). Then, <CDG=180-ADC-ADE=180-ADC-ACD=CAD, so DG and analagously CG are tangent lines to (ACD). Now, it's clear that AG is a symmedian of ACD, whence AF is the median isogonal to it. $\blacksquare$

Took me some time to figure it out... I started by thinking that maybe Trigonometric form of Ceva's Theorem works... But simple stuff worked out. Solution: One can easily see that $ABCD \sim ACDE$. This would give us the ratio that \[\frac{AY}{YC} = \frac{AX}{XC}.\]One application of Ceva's Theorem in $\triangle ADC$ at point $P$ would now solve the problem. $\blacksquare$

Let $BD\cap AC=X$ and let $CE\cap AD=Y$. Notice that $ABCDX\sim ACDEY$ and so $\frac{AX}{XC}=\frac{AY}{YD}$. Finish by Ceva's.

Draw initially the points $A, B, C$ counter-clockwise, defining the rest, clearly, counter-clockwise too. Notice that $CD$ is tangent to $(ABC)$ and $(AED)$, because

Now, let's redefine $P$. Let $P\neq A$ the second intersection of $(ABC)$ and $(ADE)$.

Finally, we can easily prove that $C, P, E$ are collinear (similar for $D,P, B$), this happens because $\angle APC= \angle BAC+\angle ACB$ and $\angle EPA= \angle ADE$ and look: $\angle BAC+\angle ACB+\angle ADE= \angle BAC+\angle ACB+\angle ABC= 180$. We can finish from here, let $M= AP\cap CD$, so $M$ lies in the radical axes of $(ABC)= \omega_1$ and $(ADE)=\omega_2$ $Pow_{\omega_1}M= MC^2= MP\cdot MA$ $Pow_{\omega_2}M= MD^2=MP\cdot MA$ $\implies MC^2=MD^2\implies MC=MD$, as desired.

Let $\overline{BD}$ intersect $\overline{AC}$ and $F$, let $\overline{CE}$ intersect $\overline{AD}$ at $G$. Since quadrilateral $ABCD$ is similar to quadrilateral $ACDE$, it follows that $\frac{AF}{AC} = \frac{AG}{AD}$, hence by Ceva's theorem, $\overline{AP}$ bisects $\overline{CD}$.

Let $Q$ and $R$ be the intersections of $AC$ with $BD$ and $AD$ with $CE$ respectively. As quadrilaterals $ABCD$ and $ACDE$ are similar, $\frac{AQ}{QC} = \frac{AR}{RD}$. The result follows as a direct consequence of Ceva's theorem.

$\color{magenta}\boxed{\textbf{SOLUTION G3}}$ $\triangle ABC$ ∼ $\triangle ADE$ $\implies A$ is the center of the spiral similarity $S : BC \to DE.$ Hence $BD \cap CE \equiv P \implies A$ lies on the circumcircles of $\triangle PBC$ and $\triangle PDE$ that is $ABCP$ and $APDE$ are cyclic. $\angle ABC = \angle ACD \implies CD$ is tangent to $(ABC)$ Similarly, $CD$ is tangent to $(ADE)$ Let $M\equiv AP \cap CD$ By $\textbf{POP,}$ $$MD^2=MP.MA=MC^2 \implies MD=MC \blacksquare$$

Let $X = AD \cap CE$ and $Y = AC \cap BD$. From given angle conditions, we obtain pairwise similar triangles $$\Delta ADE \sim \Delta ACD \sim \Delta ABC$$and it follows that $$\dfrac{DE}{DC} = \dfrac{AD}{AC} = \dfrac{DC}{BC}$$Further, we have $$\angle EDC = \angle EDA + \angle ADC = \angle DCA + \angle ACB = \angle DCB$$and this leads to $\Delta EDC \sim \Delta DCB$ by side-angle-side similarity. Angle chasing thus gives \begin{align*} \angle CED &= \angle BDC \\ \angle CED + \angle EDA &= \angle BDC + \angle DCA \\ \angle EXA &= \angle DYA \end{align*}Since it is given that $\angle EAD = \angle DAC$, we have $\Delta AYD \sim \Delta AXE$, which yields $$\dfrac{AY}{AX} = \dfrac{AD}{AE} = \dfrac{AC}{AD}$$so $\Delta AXY \sim \Delta ADC$ by side-angle-side similarity, and $XY \parallel DC$. Letting $M = AP \cap DC$, the result follows by Ceva's on the concurrence point $P$ in $\Delta ADC$. $\blacksquare$

We are basically given that $\triangle ABC \sim \triangle ACD \sim \triangle ADE$ Claim: $ABCP$ and $APDE$ are cyclic. Proof: Let $(ABC) \cap (ADE)=P'$, now by angle chasing: $$\angle AP'E=\angle ADE=\angle ABC=180-\angle AP'C \implies C,P',E \; \text{colinear!}$$In the same way we can prove $B,P',D$ colinear therefore $P=P'$ and the claim is true. Finishing: Because of the angles given by the similarities we have that $(ABC), (ADE)$ have $CD$ as one of their common tangents, let $M$ the midpoint of $CD$ then as $M$ has the same power of point on both circles we have $A,P,M$ colinear by radical axis, thus we are done

Sketch Observe that $(ABC)$ and $(ADE)$ are tangent to $CD$, now angle chase gives $P=(ABC)\cap(ADE)$ Power of point gives that AP is the median of $\Delta ACD$ so we are done

Prove that by Spiral Similarity $EDPA$ and $BCPA$ are cyclic.By the easy angle chasing we can show that $(DC)$ is common tangent of $(EDPA)$ and $(BCPA)$. $DC$ is common tangent and $AP$ is radical axis by the Power we can show that $AP$ bisects $DC$.

By the given angle conditions, we see that $A$ is the center of spiral similiarity sending $BC \rightarrow DE$ so $A$ is also the center of spiral similarity sending $BD \rightarrow CE$ so this implies that $PAED$ and $PABC$ are cyclic quadrilaterals due to the well known spiral similarity lemma. Now some angle chasing gives us that $CD$ is tangent to the circumcircles of both of these cyclic quadrilaterals and this finishes since their radical axis is $AP$ and it is well known that the radical axis passes through the midpoint of the common tangent.

Problème très intéressant! Denote by $M$ the intersection of line $\overline{AP}$ with side $CD$. The given angle condition basically tells us that $\triangle ABC \sim \triangle ACD \sim \triangle ADE$. We first make the following simple observation. Claim : Triangles $\triangle BCD$ and $\triangle CDE$ are similar. Proof : First note that, \[\measuredangle DCB = \measuredangle DCA + \measuredangle ACB = \measuredangle EDA + \measuredangle ADC = \measuredangle EDC \]Further since $\triangle ABC \sim \triangle ACD \sim \triangle ADE$ we also have, \[\frac{BC}{CD}=\frac{AC}{AD}=\frac{CD}{DE}\]which implies that $\triangle BCD \sim \triangle CDE$ as desired. Now we can attack the bulk of the problem. The following is the key claim. Claim : Quadrilaterals $AEDP$ and $ABCP$ are cyclic. Proof : This is a simple angle chase. Note that, \[\measuredangle DPE = \measuredangle DEP + \measuredangle PDE = \measuredangle DEC + \measuredangle BDE = \measuredangle DEC + \measuredangle CDE + \measuredangle BDC = \measuredangle CDE = \measuredangle DAE \]which indeed implies that $APDE$ is cyclic. A similar argument also shows that $ABCP$ is cyclic, proving the claim. Further note that, \[\measuredangle PDC = \measuredangle BDC = \measuredangle CED = \measuredangle PED\]so $\overline{CD}$ is actually tangent to $(APDE)$ at $D$. Similarly, $\overline{CD}$ is also tangent to $(ABCP)$ at $C$. Now, it is well known that the common external tangent to two intersecting circles, is bisected by their radical axis. Thus, considering circles $(APDE)$ and $(APCB)$ it follows that line $AP$ bisects side $CD$ as desired.

Note that $\triangle ABC \stackrel{+}{\sim} \triangle ACD \stackrel{+}{\sim} \triangle ADE$. Now $Q = BC \cap DE$ is the pole of $CD$ with respect to $(ACD)$ since \begin{align*}\angle QCD = 180^\circ - \angle ACB - \angle ACD = 180^\circ - \angle ADC - \angle ACD = \angle CAD\end{align*}and \begin{align*}\angle QDC = 180^\circ - \angle ADE - \angle ADC = 180^\circ - \angle ACD - \angle ADC = \angle CAD.\end{align*}Hence $AP$ and $AQ$ are isogonal in $\angle CAD$ by DDIT on $BCED$ at $A$; since $AQ$ is the $A$-symmedian we're done.