in two cyclic quadrilatrals APRD & DRCQ ,AD & CD are diameters respectively and we have: RQ/sin(<RCQ)=CD/2 or RQ/sin(<BCA)=CD/2 PR/sin(<PAR)=AD/2 or PR/sin(<BAC)=AD/2 By dividing: CD/AD= sin(<BAC)/sin(<BCA)=BC/AB Since CD/AD=BC/AB thus bisectors of <CBA and <ADC intersect AC in the same poin. S.Ebadollahi

i must admit that this problem was quite easy by imo standards.my solution is based on a standard trick using pedal triangles and projections which is damn easy and pretty similar to the proof posted here.

here is an interesting thing about this problem and other 2 IMO problems in the past! it's about the so called Pedal Triangle Trick! this was posted on a forum by: fritue2000 and I thought it would be interesting to post it here also! "The previous two IMO problems solved by the Pedal Triangle Trick. (1996 IMO) Let P be a point inside the triangle ABC such that angle APB - angle ACB = angle APC - angle ABC. Let D, E be the incenters of triangles APB, APC respectively. Show that AP, BD, CE meet at a point. As I know, at the jury meeting at IMO 1996, there is a discussion for this problem becasuse its solution is very similar to problem 2 of the 34th IMO. The techniques include 'inversion' and 'PTT', etc. For example, http://home1.pacific.net.sg/~slwee/imo96/imo96op.htm ) (1993 IMO) Let D be a point inside the acute-angled triangle ABC such that angle ADB = angle ACB + 90 degrees, and AC*BD = AD*BC. (a) Calculate the ratio AB*CD/(AC*BD). The well-known 'Pedal Triangle Trick' is "For any point D, let X, Y, Z be feet of the altitudes from D to AB, BC and CA. Then, YZ = (DA*BC)/2r, etc, where r is the circumradius of ABC." The proof is very easy, since D, A, Y, Z lies on a circle with diameter DA, by the law of sines, YZ = DA sin A = DA*(BC/2r). (2003 IMO) Given is a cyclic quadrilateral ABCD and let P, Q, R be feet of the altitudes from D to AB, BC and CA respectively. Prove that if PR = RQ then the interior angle bisectors of the angles <ABC and <ADC are concurrent on AC. Solution) By PTT, PR=RQ implies (DA*BC)/2r = (DC/AB)/2r, so, CD/DA=BC/AB implies the results. q.e.d. The most noticeable thing is in the above solution, we did not used the condtion 'ABCD is cyclic'. And as an IMO problem, it is not so intersting because all three problems 1993, 1996, 2003 solved by exactly the same TWO ways (PTT and inversion). Well, the PTT is really well-known, for example, it appears at "Geometry Revisited" by Coexter. It also used for a proof of Ptolemy's Theorem. In a cyclic qudrilateral ABCD, the three points are on the simson line, we have PR+RQ=RP, with the same notation of 2003 imo. Then, (DA*BC)/2r + (DC*AB)/2r = (DB*AC)/2r or DA*BC + DC*AB = DB*AC. "

A bit of standard angle chasing shows that triangles DPR and DBC are similar and so are DQR and DBA. Thus PR=QR gives DC/CB = DR/RP = DR/RQ = DA/AB, and the result follows from the angle bisector theorem.

Yeah, I had something like that too, was quite easy solved in about 30 minutes (which is extremely few for me)

Lemma: Let $ABCD$ be a cyclic quadrilateral. Let $DR$ and $DP$ be the altitudes from $D$ to $AB$ and $DC$, respectively. Let $RP$ intersect $AC$ at $Q$. Then $\angle AQD$ is a right angle. Proof of Lemma: $\angle DRB = \angle DPB = \pi / 2$, so $DRPB$ is cyclic. Then $\angle DRP = \angle DBP$. From cyclic quadrilateral $ABCD$, $\angle DBP = \angle DAP$. $\angle DRP = \angle DAP$, so $RAQD$ is cyclic. Thus $\angle AQD = \pi - \angle ARD = \pi / 2$. Proof: Let the bisector of $\angle ABC$ meet $AC$ at S. Let the bisector of $\angle ADC$ meet $AC$ at T. By the Lemma, $R,Q,P$ are collinear. Let $\angle DRQ = \alpha$. $\angle APD = \pi - \alpha - \angle RDP = \pi - \alpha - (\pi - \angle ABC) = \angle ABC - \alpha$. $\frac{RQ}{\sin \angle RDQ} = \frac{DQ}{\sin \alpha}$, $\frac{PQ}{\sin \angle PDQ} = \frac{DQ}{\sin (\angle ABC - \alpha)}$. $\frac{RQ}{PQ} \cdot \frac{\sin PDQ}{\sin RDQ} = \frac{sin(\angle ABC - \alpha)}{\sin \alpha}$. $\angle RDQ = \pi - \angle RAQ = \angle BAC$. Also $\angle PDQ = \angle BCA$. $\frac{RQ}{PQ} \cdot \frac{\sin \angle BCA}{\sin \angle BAC} = \frac{\sin (\angle ABC - \alpha)}{\sin \alpha}$. $\frac{RQ}{PQ} \cdot \frac{AB}{BC} = \frac{\sin \angle QCD}{\sin \angle QAD} = \frac{AD}{DC}$. So $RQ=PQ$ if and only if $\frac{AB}{BC} = \frac{AD}{DC}$. By the Angle Bisector Theorem, $\frac{AS}{SC} = \frac{AB}{BC}$ and $\frac{AT}{TC} = \frac{AD}{DC}$. So $\frac{AB}{BC} = \frac{AD}{DC}$ if and only if $\frac{AS}{SC} = \frac{AT}{TC}$. This can happen if and only if $S=T$. So $PQ=QR$ if and only if the angle bisectors of $\angle ABC$ and $\angle ADC$ are concurrent with $AC$. QED.

Yay I exploded this problem with like 20 cyclic quadrilaterals. Okay first well-know fact: Simson's Line, so we know P, Q, R are collinear. i'll go with the only if direction Let X be the intersection of the angle bisectors. Extend BX, DX to Y, Z, where Y and Z lies on the circumcircle of ABCD. Y, Z lies on the perpendicular bisector of AC, hence YZ is a diameter, and it goes through midpoint M of AC. angle AMY = 90, angle ZDY = 90 because YZ is a diameter, so MYDX is cyclic, so angle ZYB = MDX = ADB. DX bisects ADC hence ADM = BDC = BAC = PDQ. angle QAD=QPD. Hence triangle PQD is similar to triangle AMD, and triangle ADC is simlar to PDR. DM is a median hence DQ is a median => Q is the midpoint of PR as desired. If direction: Define X to be on AC such that DX bisects ADC, DX intersect the circumcircle at Z, M is the midpoint of AC, Y is the intersection of MZ and the circumcircle. triangle ADC is simlar to PDR. DM is a median and DQ is a median, so triangle PQD is similar to triangle AMD, angle ADM = PDQ=BAC = BDC. By definition XD bisects ADC hence XD bisects MDB so MDX = XDQ. QMY=90, YZ is still diameter, MYDX is cyclic, ZYX = MDX = ZDB. Now let BX intersect the circumcircle at Y'. XY'Z = XYZ, now Y' is unique(consider the locus of points Y' such that XY'Z = XYZ, which is a circle, and Y' lies on the cirucmcircle of ABCD. Two circle intersects in two points, one of which is Z, hence there's only one such point left and thus it's unique), hence Y = Y', thus B, X, Y are collinear. Y lies on the perpendicular bisector of AC so YA=YC, thus angle ABY = YBC. QED

The following will be useful: Let $ \triangle ABC$ be a triangle, $ D$ an arbitrary point and $ X$, $ Y$, $ Z$ the pedal points wrt the sides $ BC$, $ CA$, $ AB$. Then $ YZ= \frac{AD \cdot BC}{2R}$, where $ R$ denotes the circrumradius of $ \triangle ABC$. That's very easy to prove, I won't write it down now . Using the lemma shown above, we have $ PQ= \frac{CD \cdot AB}{2R}$ and $ QR= \frac{AD \cdot BC}{2R}$. Hence, $ PQ =QR \Leftrightarrow \frac{AB}{BC} = \frac{AD}{DC}$. Let $ S$ denote the intersection of the angle bisectors of $ \angle ABC$ and $ \angle CDA$. If $ S \in AC$, then $ \frac{AB}{BC} = \frac{AS}{SC} = \frac{AD}{DC}$. If $ PQ=QR$, then $ \frac{AS_1}{S_1C} = \frac{AB}{BC} = \frac{AD}{DC} = \frac{AS_2}{S_2C}$, hence $ S_2=S_1=S$, where $ S_1$ and $ S_2$ denote the intersections of the angle bisectors of angles $ \angle ABC$ and $ \angle CDA$ with the side $ AC$. Thus, the problem is proved.

Hmmm I came up with this, which uses a bit more of Simson line properties. Let the angle bisector of $ \angle{ABC}$ intersect $ AC$ at $ N$, let the midpoint of $ AC$ be $ M$, let $ E$ and $ F$ be points diametrically opposite on the circumcircle of $ ABCD$ such that $ EF$ is the perpendicular bisector of $ AC$, with $ E$ on the same side of $ AC$ as $ B$. Let $ B'$ be the point diametrically opposite to $ B$. $ FDQN$ is cyclic, $ FDMN$ is a rectangle (cyclic), so $ \angle{FND} = \angle{FQD} = \angle{MDQ}$. $ \angle{FND} = \angle{FBD} + \angle{BFE}$ $ \angle{B'BF} = \angle{BFE}$, since both are diameters. Thus $ \angle{B'BD} = \angle{BNE} = \angle{MDQ}$. $ P,Q,R$ are collinear and are the Simson line of $ D$. $ AC$ is the Simson line of $ B'$. By a well known property of Simson lines, $ \angle{PQA} = \frac12\overarc{B'D}$, but $ \frac12\overarc{B'D} = \angle{B'BD} = \angle{BNE} = \angle{MDQ}$ and $ \angle{PQA} = \angle{PDA} = \angle{CDR}$, since $ PDR$ is directly similar to $ ADC$. Thus we have a spiral similarity with center $ D$ mapping $ A,M,C$ to $ P,Q,R$ respectively, and since $ M$ is the midpoint of $ AC$, $ Q$ is the midpoint of $ PR$ so we are done. The argument should work in reverse as well, with some tweaking here and there.

The concurrent with $AC$ bisectors condition is equivalent to $\frac{AB}{BC}=\frac{AD}{DC}$ $=> \frac{AD*BC}{AB*DC} =1$ Since $\angle DAB+ \angle DCB= 180$ and $\angle DCP+ \angle DCB= 180$ we have $\angle DAB= \angle DCP$. Therefore we have that $ARD$ is similar to $CPD$ and therefore $\frac{AR}{PC}=\frac{AD}{DC}$. (1) By Menelaus theorem for transversal $CQA$ in triangle $BRP$ we obtain: $\frac{AB*RQ*PC}{AR*PQ*BC}=1$. Therefore $\frac{RQ}{QP}=\frac{AR*BC}{AB*PC}=\frac{AD*BC}{AB*DC}$ (because of (1)). The conclusion is now obvious.

Use that a quadrilateral is harmonic if and only if the diagonals are symmedians(easy to prove with polarity and crossed ratio). Now, the problem is equivalent to this: $\triangle{ABC}$,$D$ is on the circumcircle of $\triangle{ABC}$(on the arc$BC$). Denote $X,Y,Z$ the projections of $D$ onto $AB,BC,CA$. $XY=YZ$if and only if $D$ is on the symmedian with respect to $A$. Proof: See that $BXDY$ and $DYZC$ are inscribed in circles with diameter$BD$,respectively$CD$. We have from Simson's line that $X-Y-Z$are collinear. In $\triangle{XBY}$(aplying law of sines) $XY=BDsinB$, and in $\triangle{YZC}$ we have$YZ=DCsinC$. So, $XY=YZ$ if and only if $\frac{BD}{CD}=\frac{sinC}{sinB}$, but it happens only for symmedian(note that $\frac{sinBAM}{sinCAM}=\frac{sinB}{sinC}$, where $M$ is the midpoint of $BC$.

hey guys) I solved this problem in about 10 minutes, but I think there is some thing wrong because I didn't use the fact that $ABCD$ is cyclic. Can it be so?

A different approach: To begin with, it is clear that triangles $\bigtriangleup DRP$ and $\bigtriangleup DAC$ similar. (This follows by a very simple angle chasing, as in the above posts). This also gives us the fact that triangles $\bigtriangleup DAR$ and $\bigtriangleup DPC$ are similar, therefore $\frac{DA}{DC}=\frac{RA}{PC} (1)$ Also, points $P,Q,R$ are collinear (Simson's line) It remains to prove the equivalence $PQ=QR \Leftrightarrow \frac{BA}{BC}=\frac{DA}{DC}$ Let's focus on triangle $\bigtriangleup RBP$ . Points $A,Q,C $ are collinear and lie on the lines $BR,RP,PB$ respectively. Applying Menelaus' Theorem, we have: $\frac{AR}{AB}\cdot \frac{CB}{CP}\cdot \frac{QP}{QR}=1$ Therefore, $QP=QR\Leftrightarrow \frac{QP}{QR}=1\Leftrightarrow \frac{AB}{BC}=\frac{RA}{PC}\Leftrightarrow \frac{AB}{BC}= \frac{DA}{DC}$, since $\frac{DA}{DC}=\frac{RA}{PC}$ by relation (1). The conclusion follows immediately. Nick

nickthegreek wrote: A different approach: Your solution is actually identical to hatchguy's. Take a look: hatchguy wrote: The concurrent with $AC$ bisectors condition is equivalent to $\frac{AB}{BC}=\frac{AD}{DC}$ $=> \frac{AD*BC}{AB*DC} =1$ Since $\angle DAB+ \angle DCB= 180$ and $\angle DCP+ \angle DCB= 180$ we have $\angle DAB= \angle DCP$. Therefore we have that $ARD$ is similar to $CPD$ and therefore $\frac{AR}{PC}=\frac{AD}{DC}$. (1) By Menelaus theorem for transversal $CQA$ in triangle $BRP$ we obtain: $\frac{AB*RQ*PC}{AR*PQ*BC}=1$. Therefore $\frac{RQ}{QP}=\frac{AR*BC}{AB*PC}=\frac{AD*BC}{AB*DC}$ (because of (1)). The conclusion is now obvious.

"Futurama" explains the collinearity of P, Q and R. ADC and PDR are similars, then we know that AD/DC=AR/PC Let P' a point on BC such that PC=P'C. RQ=QR iff The altitudes from P and R to AC are equals iff P'R//AC iff AB/BC=AR/P'C=AR/PC=AD/DC and finish

by simson's theorem we know $ P,Q,R $ are co-liner . at first assume $RQ=QP$ $\triangle DRP \sim \triangle DAC$ $RDQA$ is cyclic so $RDQ=CAB$ $DCPQ$ is cyclic so $QDP=ACB$ then we have $\frac{DR}{DP}=\frac{sinPDQ}{sinRDQ} = \frac {DA}{DC}= \frac {sinCAB}{sinACB}=\frac {AB}{BC}$ and do this back

Here's a basic projective approach: Let E,F be the intersection points of the line through B parallel to PR and BQ, each with the circumcircle, respectively. Notice from cyclic quad CDQP and PQR being a simson line, we get $QDC=180-QPC=BPQ=EBC=EDC$, which implies that EQD is a line. Now, $$-1=(PR;QP_{\infty})=^B(CA;FE)=^Q(AC;BD)$$harmonic, as desired. $\blacksquare$

Note that $PQR$ is a Simson Line. Let $PD$ intersect $(ABCD)$ again at $E$. Note that $AE \parallel PQ$ (this is well known, just an angle chase). Now, let the point at infinity of lines $PQ$ and $AE$ be $\infty$. Then \[ (P,R; Q, \infty) \stackrel{A}{=} (P, B; C, (AE \cap BC)) \stackrel{E}{=} (D, B; C, A) \]which implies that $Q$ is the midpoint of $PR$ if and only if $ABCD$ is harmonic, if and only if the bisectors of $\angle ABC$ and $\angle ADC$ intersect on $AC$. $\blacksquare$

Let $E$ be the intersection of the bisector of $\angle ADC$ and $AC$.The complete diagram is shown below. We know that, $P$, $Q$ and $R$ are colinear by Simson's Line. We have the following quadrilaterals cyclic : \[\angle ARD=\angle AQD=90^{\circ}\implies AQDR\ cyclic\]\[\angle BRD=\angle BPD=90^{\circ}\implies BPDR\ cyclic\]\[\angle CQD=\angle CPD=90^{\circ}\implies CDQP\ cyclic\]Angle chasing gives us, $\angle DQR=\angle DAR=\angle BCD$ and $\angle QRD = \angle PRD=\angle PBD=\angle CBD$ Which implies, $\triangle QRD\sim \triangle BCD\implies \frac{QR}{DQ}=\frac{BC}{CD}$ Again, $\angle DBA=\angle DBR=\angle DPR=\angle DPQ$ and $\angle BAD=180^{\circ}-\angle DAR = 180^{\circ}-\angle DQR=\angle PQD$ Which implies, $\triangle ABD\sim \triangle PQD\implies \frac{PQ}{DQ}=\frac{AB}{AD}$ Now, suppose $E'$ is the intersection of the $AC$ and bisector of $\angle ABC$. By Angle Bisector Theorem, we know that, $\frac{AD}{CD}=\frac{AE}{EC}$ and $\frac{AB}{BC}=\frac{AE'}{E'C}$. So, The angle bisectors of $\angle ABC$ and $\angle ADC $ concur if and only if $E=E'$ \[\frac{AE}{EC}=\frac{AE'}{E'C}\iff \frac{AD}{CD}=\frac{AB}{BC}\iff\frac{AB}{AD}=\frac{BC}{CD}\iff\frac{PQ}{DQ}=\frac{QR}{DQ}\iff PQ=QR \]as desired. $\Box$

here's a fourth projective solution to this i guess Note that $\overline{PQR}$ is the Simson line of $D$ with respect to triangle $ABC$. Hence for $E = \overline{DP} \cap (ABCD)$, it is parallel to $\overline{AE}$. Then $$(AC;BD) \stackrel P=(FB;EC) \stackrel A= (PR; Q\infty_{PQR})$$which implies the result.

Notice that the concurrent angle condition is identical to the condition that $ABCD$ is a harmonic quadrilateral. Observe now that \[\frac{PQ}{QD} = \frac{\sin \angle PDQ}{\sin \angle QPD} = \frac{\sin \angle BAC}{\sin \angle DAC},\]and \[\frac{QR}{QD} = \frac{\sin \angle QDR}{\sin \angle DRQ} = \frac{\sin \angle ACB}{\sin \angle DBA}. \]Hence, \[\frac{PQ}{QR} = \frac{\sin \angle BAC}{\sin \angle DAC} \cdot \frac{\sin \angle DBA}{\sin \angle ACB} = \frac{BC}{BA} \cdot \frac{DA}{DC} = 1 \iff ABCD \text{ is harmonic,}\]as desired. $\blacksquare$

If the bisectors meet on AC this gives us $\frac{AD}{DC} = \frac{AB}{BC}$. From ABCD cyclic $\angle BAC = \angle PCD$, so $\triangle CPD \sim \triangle ARD$, so $\frac{AR}{CP} = \frac{AD}{DC}$. From Menelaus on $\triangle BPR$ we get $\frac{BA}{AR}. \frac{RQ}{QP}. \frac{PC}{CB} = 1$ (we can do this since P, Q, R lie on one line aka Simpson line) and $\frac{BA}{AR}. \frac{PC}{CB} = \frac{AD}{DC}. \frac{DC}{AD} = 1$. This means $\frac{PQ}{QP} = 1$ $\Rightarrow$ PQ = QP. In the other direction is the same because the pair of similar triangles is there, so we get that $\frac{BA}{CB} = \frac{AD}{DC}$ which is equivalent to the fact that the bisectors meet on AC. We are ready.

Notice that $P,Q,R$ lie on the Simson line. Let $X$ be the second intersection of line $DQ$ with the circle $\omega$ circumscribing $ABCD$. We claim that $\overline{BX} \parallel \overline{RQ}$. Using directed angles, \[ \angle BXD = \angle BAD = \angle RAD = \angle RQD, \]and thus, $\overline{BX}$ is parallel to the Simson line. Next, note that by the angle bisector theorem, the bisector condition is equivalent to $ABCD$ being harmonic. Additionally, we have \[ (A,C;B,D) \stackrel{X}{=} (A,C;\overline{BX} \cap \overline{AC}, Q) \stackrel{B}{=} (R,P;P_\infty,Q), \]and since $(R,P;P_\infty,Q) = -1 \iff PQ=QR$, we are done.

Complex bash, let $A = w, C = \overline{w}, B = b, D = d$. Let $M,N = 1,-1$ be the arc midpoints of $AC$. Note $BD,MN$ meet a point whose polar is parallel to $AC$. If $BM,DN,AC$ are concurrent (or $BN, DM, AC$), then Brokard tells us that $AC$ is precisely this polar, so the condition can be reduces to the intersection of tangents from $A,C$ being the intersection of $BD, MN$. The intersection of $BD, MN$ is just $\frac{b + d}{bd + 1}$, so the desired condition is just $\frac{b + d}{bd + 1} = \frac{2}{w + \overline{w}}$, rearrange to get $2bd + 2 = (b + d)(w + \overline{w})$. Now for the condition that $PQ = QR$. We dilate by two and instead find a condition for $P'Q' = Q'R'$. By Simson line, it just suffices to show $p + r = 2q$. The reflection of $D$ over $AB$ is just $w + b - \frac{bw}{d}$, likewise the reflection of $D$ over $BC$ is just $b + \overline{w} - \frac{b\overline{w}}{d}$. The reflection of $D$ over $AC$ is just $w + \overline w - \frac 1d$. Now the condition is $2b + w + \overline{w} - \frac bd (w + \overline{w}) = 2(w + \overline{w} ) - \frac 2d$. Multiplying by $d$, an equivalent condition is $2bd + (d -b)(w + \overline{w}) = 2d(w + \overline{w}) - 2$, rearranging we get $(b + d)(w + \overline{w}) = 2bd + 2$, as desired. Since the conditions are equivalent, we can conclude that $PQ = QR$ and the concurrency happen at exactly the same configurations, we are done.

First, note that if the angle bisectors of $\angle ABC$ and $\angle ADC$ meet on segment $AC$, we then have \[\frac{AB}{CB}=\frac{AD}{CD} \iff AB*CD=AD*BC \iff (AC;BD)=-1,\]and by the uniqueness of the harmonic conjugate, this means that $B$ is the unique point on $(ACD)$ such that $(AC;BD)=-1$. By the harmonic quadrilateral configuration, this means that $BD$ is the $D$-symmedian of $\triangle ACD$. It then suffices to show that \[PQ=QR\iff BD\text{ is the } D\text{-symmedian of }\triangle ACD.\] To start, we make the following claims. *** Claim 1. $P$, $Q$, $R$ are collinear. Proof. This is immediate from the Simson Line at $D$ with respect to $\triangle ABC$. *** Claim 2. There exists a spiral similarity centered at $D$ sending $\triangle DRP$ to $\triangle DAC$. Proof. All angles below are directed. We can prove this through (directed) angle chasing. First, note that since $\angle CQD=\angle CPD=90$, we have that $C$, $P$, $D$, $Q$ are concyclic, which gives us that \[\angle RPD=\angle QPD=\angle QCD=\angle ACD,\]and since $\angle ARD=\angle AQD=90$, we have that $A$, $R$, $Q$, $D$ concyclic, meaning that \[\angle PRD=\angle QRD=\angle QAD=\angle CAD,\]and these two combined gives us that $\triangle DRP\sim \triangle DAC$, as desired. This proves our claim. *** Now, let $M$ be the midpoint of $AC$. Notice that $Q$ is the midpoint of $RP$ if and only if the spiral similarity described in Claim 2. sends $Q$ to $M$. However, note that \[\angle Q\rightarrow M \iff \angle MDQ=\angle CDP \iff 90-\angle MDQ=90-\angle CDP \iff \angle DMQ=\angle DCP,\]and since $\angle DCP=\angle BAD$ by cyclic properties, and $\angle DMQ=\angle DMC$, this gives us that \[\iff \angle BAD=\angle DMC.\] Now, let us fix $A$, $C$, and $D$. We make the following claim. *** Claim 3. If $\angle BAD=\angle DMC$, then $B\neq D$ must be the unique point on $(ACD)$ such that $BD$ is the $D$-symmedian of $\triangle ACD$. Proof. Notice that since $\angle DMC$ is fixed, $\angle BAD$ must also be fixed, meaning that there is a unique point on $(ACD)$ such that $\angle BAD=\angle DMC$. Now, we can angle chase. We get that \[\angle ADM=180-\angle AMD-\angle MAD=\angle CMD-\angle MAD=\angle BAD-\angle MAD=\angle BAC=\angle BDC,\]which means that $BD$ is indeed the $D$-symmedian of $\triangle ABC$, as desired. This proves our claim. *** By Claim 3. means that $\angle BAD=\angle DMC$ is equivalent to the condition that $BD$ is the $D$-symmedian of $ADC$. However, we also established earlier that the former was equivalent to $Q$ being the midpoint of $PR$, meaning that \[PQ=QR\iff BD\text{ is the } D\text{-symmedian of }\triangle ACD,\]as desired. Since \[BD\text{ is the } D\text{-symmedian of }\triangle ACD\iff (AC;BD)=-1\iff \text{the angle bisector of }\angle ABC \text{ and } \angle ADC \text{ meet on } AC,\]we have that the angle bisectors meet on $AC$ if and only if $PQ=QR$, which is what we wished to prove. This completes our proof.

Notice that \[PQ = QR \iff AD \sin \angle A = CD \sin \angle C \iff AD \cdot BC = CD \cdot AB. \quad \blacksquare\]

Let $DP$ intersect $(ABC)$ again at $K$. By Simson line $P, Q, R$ are collinear, and $\measuredangle QPD = \measuredangle PCD = \measuredangle AKD$ so $AK \parallel PQ$. Then $$(AC;BD) \overset{P}{=} (AP \cap (ABC), B; C, K) \overset{A}{=} (PR; Q\infty),$$which concludes by angle bisector theorem.

arshakus wrote: hey guys) I solved this problem in about 10 minutes, but I think there is some thing wrong because I didn't use the fact that $ABCD$ is cyclic. Can it be so? Yes. (same here) it is not a requirement at all . The fact that it is cyclic helps in the creation of simson line and just makes the diagram easy to construct plus it gets rid of nasty configuration issues. I was able to see some possible issues(which were not quite problematic are so) But there could have been some config that I missed. I asked a few people too and all of them agree. IMO committee wanted the problem to own its own merit instead of relying on configs so yippee

$\boxed{nice problem}$

How can it be to IMO? \[PQ = QR \iff AD \sin \angle A = CD \sin \angle C \iff AD \cdot BC = CD \cdot AB. \quad \blacksquare\]

My first IMO #4 sol We notice that second condition is equivalent to showing that $ABCD$ is an harmonic quadrilateral because of the angle bisector theorem, so we want to proof that $\frac{CB}{AB}=\frac{CD}{AD}$. Assuming that $PQ=QR$, it's well known that $P-Q-R$, so we proceed by applying Menelaus's theorem on $\triangle PRB$ with transversal line $AC$, getting that $\frac{CB}{AB}=\frac{CP}{AR}$, so now we want to show that $\frac{CP}{AR}=\frac{CD}{AD}$, but just notice that $\triangle DPC \sim \triangle DRA$ which gives the proportion. Assuming that $\frac{CB}{AB}=\frac{CD}{AD}$, by the previous proportion we know $\frac{CP}{AR}=\frac{CD}{AD}$ so by applying Menelaus's theorem again on $\triangle PRB$ with transversal line $AC$, we obtain $PQ=QR$, so we're done. $\square$