My solution: Let $ P^*=PD \cap AB, Q^*=QD \cap AC $ and $ AB=AC=\delta $ . Easy to see $ P, Q $ is the midpoint of arc $ AB $ in $ \odot (ABD), $ arc $ AC $ in $ \odot (ACD) $, respectively . From the condition $ \Longrightarrow \frac{BD}{AD}+\frac{CD}{AD}=1 \Longrightarrow \frac{BP^*}{\delta + BP^*}+\frac{CQ^*}{\delta + CQ^*}=1 \Longrightarrow \delta^2=BP^* \cdot CQ^* $ , so from $ \frac{P^*B}{P^*A}=\frac{DB}{DA} \Longrightarrow P^*B \cdot DA=P^*A \cdot DB = \delta \cdot DB+P^*B \cdot DB $ $ \Longrightarrow P^*B \cdot DC= \delta \cdot DB \Longrightarrow ( P^*B \cdot DC )^2=\delta ^2 \cdot DB^2=P^*B \cdot Q^*C \cdot DB^2 \Longrightarrow \frac{DC^2}{Q^*C}=\frac{DB^2}{P^*B} $ . ... $ (\star) $ If we denote $ R_1=BP \cap AD, R_2=CQ \cap AD $ , then from $ (\star) $ we get $ \frac{AR_1}{DR_1}=\frac{[ABP]}{[DBP]}=\frac{ \delta \cdot AP}{DB \cdot DP}=\frac{ \delta \cdot BP^*}{DB^2}=\frac{ \delta \cdot CQ^*}{DC^2}=\frac{\delta \cdot AQ}{DC \cdot DQ}=\frac{[ACQ]}{[DCQ]}=\frac{AR_2}{DR_2} \Longrightarrow R_1 \equiv R_2 \equiv R $ , so we get $ BR\cdot RP=AR \cdot RD=CR \cdot RQ \Longrightarrow B, C, P, Q $ are concyclic . Q.E.D

Trivial problem for China Solution: apply an inversion with center $D$ and arbitrary power $K$. Let $X'$ inverse of $X$ note that points $Q$ and $P$ are midpoints of arcs $ADC$ and $ADB$ of triangles $ADC$ and $ADB$ so $Q$' and $P'$ are feet of external angle bisectors of $A'DC'$ and $A'DB'$. also from the condition we have $1/DA'=1/DB'+1/DC'$ using this fact and external bisector theorem in triangles $A'DB'$ and $A'DC'$ with easy calculation we get $A'B'/A'P'=A'Q'/A'C'$ so $B'Q'||C'P'$ and quadrilateral $B'Q'P'C'$ is trapezoid so to prove the problem we have to show that B'A'=Q'A' from the condition that $AB=AC$ and external bisector theorem we get $DC'/DB'$=$Q'C'/Q'A'$×$P'A'/P'B'$=$A'C'/A'B'$ using this fact and $Q'A'B'\sim P'A'C'$ after an easy calculation we get the result that $A'Q'=A'B'$.

It is clear that $ P $ is the midpoint of arc $ \overarc{BDA} $ of the circumcircle of $ \triangle{BDA} $ and similarly $ Q $ is the midpoint of $ \overarc{CQA} $ of the circumcircle of $ \triangle{CQA}. $ Consider the inversion with center $ D $ and arbitrary radius. I will denote inverted points by their original letter (hopefully the use will be clear). Because of the first observation, we have that $ C, A, Q $ and $ B, A, P $ are collinear (in that order). Now we proceed with four metric observations: $ (1): \frac{1}{AD} = \frac{1}{BD} + \frac{1}{CD} $ - this follows from the given condition $ (2): \frac{AB}{AC} = \frac{BD}{CD} $ - this follows from the fact that in the original diagram, $ \triangle{ABC} $ was isosceles. $ (3): \frac{AP}{BP} = \frac{AD}{BD} $ - this follows from the fact that in the original diagram, $ \triangle{APB} $ was isosceles. $ (4): \frac{AQ}{CQ} = \frac{AD}{CD} $ - this follows from the fact that in the original diagram, $ \triangle{AQC} $ was isosceles. Now we want to show that quadrilateral $ BCPQ $ is cyclic, which is equivalent to showing that $ AC \cdot AQ = AB \cdot AP. $ But by (3) and (4) we have that $ \frac{AQ}{BP} = \frac{AB \cdot BP}{AC \cdot CQ} $ so by (2) it suffices to show that $ CQ = BP. $ But by (3) and (4) again and the facts that $ AC + AQ = CQ $ and $ AB + AP = BP $ we find that $ CQ = \frac{AB}{1 - \frac{AD}{CD}} $ and $ BP = \frac{AC}{1 - \frac{AD}{BD}} $ and now by using (1) and (2) we can simplify to obtain the desired result.

We begin with a well-known lemma that a few above have cited:

Now, applying the lemma to the problem at hand, we find that $P, Q$ are the midpoints of arcs $\widehat{ADB}, \widehat{ADC}$ in $\odot(ADB), \odot(ADC)$, respectively. We will now prove that the lines $AD, BP, CQ$ are concurrent, whence Power of a Point will finish the proof. Let $X$ be the point on $\overline{AD}$ for which $AX = DB$, and let $PB, QC$ cut $AD$ at $P', Q'$, respectively. We will show that $AP / DP' = AQ' / DQ'$, whence it will follow that $P' \equiv Q'$; but first, a few synthetic observations: Since $DX = DA - AX = DC$, it follows that $\triangle DCX$ is isoceles. Furthermore, we have $\angle CDX = \angle CDA = \angle CQA$, so by side-angle-side similarity, $\triangle CDX \sim \triangle CQA.$ By spiral similarity, we also know that $\triangle CDQ \sim \triangle CXA.$ In addition, since $PA = PB, AX = DB, \angle PAX = \angle PAD = \angle PBD$, it follows by side-angle-side similarity that $\triangle PAX \sim \triangle PBD.$ By spiral similarity, we also have $\triangle PAB \sim \triangle PXD.$ Now, we are ready: By the Ratio Lemma (Law of Sines) applied to $\triangle ADB, ADC$, it follows that \[\frac{AP'}{DP'} = \frac{BA}{BD} \cdot \frac{\sin\angle ABP}{\sin\angle DBP}, \quad \quad \quad \frac{AQ'}{DQ'} = \frac{CA}{CD} \cdot \frac{\sin\angle ACQ}{\sin\angle DCQ}.\] Keeping in mind that $BA = CA$, to prove that these two ratios are equal, we need only show that \[\frac{BD}{CD} = \frac{\sin\angle DCQ}{\sin\angle ACQ} \cdot \frac{\sin\angle ABP}{\sin\angle DBP}.\] Recalling our similar triangles in mind and applying the Ratio Lemma to $\triangle ADC$, we have \[\frac{BD}{CD} = \frac{AX}{DX} = \frac{CA}{CD} \cdot \frac{\sin\angle ACX}{\sin\angle DCX} = \frac{BA}{DX} \cdot \frac{\sin\angle DCQ}{\sin\angle ACQ}.\] Because $\triangle PAB \sim \triangle PXD$, one final calculation yields \[\frac{BA}{DX} = \frac{PB}{PD} = \frac{PA}{PD} = \frac{\sin\angle PDA}{\sin\angle PAD} = \frac{\sin\angle ABP}{\sin\angle DBP}.\] Hence, $AP' / DP' = AQ' / DQ'$, so $P' \equiv Q'$, as desired. Now, let us denote the intersection of $AD, BP, CQ$ by $Y.$ Then since $Y$ lies on the radical axis of $\odot(APDB)$ and $\odot(AQDC)$, it follows by Power of a Point that $YP \cdot YB = YQ \cdot YC.$ By Power of a Point once again, this implies that $B, C, P, Q$ are concyclic, as desired. $\square$

We easily see that $AQDC$ and $BDPA$ are cyclic. Now, pick $Y$ on $AD$ with $DY=DC$, and note the angle $DCY$ is the complement of half the angle $ADC$ by the isosceles triangle. As angle $ACQ$ is the complement of half of $AQC$ and also of $ADC$, we have that in $DCA$ that $CQ$ and $CY$ are isogonal. So if $CQ$ meets $AD$ at $X$, we have $\frac {AY} {YD}\frac {AX} {XD}=\frac {AC} {DC}^2$. So $\frac {AX} {XD}=(\frac {AC} {DC})(\frac {AC} {DC})(\frac {YD} {AY})=(\frac {AC} {DC})(\frac {AC} {DC})(\frac {DC} {DB})$, using $DA=DB+DC$. This works out to $\frac {(AB)(AC)} {(DB)(DC)}$. So then this expression is symmetric, and looking at where $BP$ meets $AD$ gives the same result. Thus $BP$, $CQ$, $AD$ meet at $X$. So then we have $(XB)(XP)=(XA)(XD)=(XQ)(XC)$ using POP and this finishes the problem.

With an inversion at D radius 1, the problem becomes: "Let 1/DC+1/DB=1/DA and AC/DC = AB/DB and P,Q be the intersections of AC, AB with the external bisector of ADC and ADB. Prove BPQC is cyclic". Notice CP/PA=CD/DA=CD*(1/DC+1/DB)=1+(CD/DB) Thus CA/PA=CD/DB and this implies PA*CA=(CA^2)/(CD/DB) and similarly QA*BA=(BA^2)(CD/DB) and thus all that remains is to prove CA^2/(CD/DB)=BA^2(CD/DB), which is equivalent to AC/DC=AB/DB, which is given in the problem statement.

First off $Q$ is the midpoint of the arc $CDA$ because is unique point that satisfies this. Let $K$ be a point on the extension of $CD$, such that $KD=DB \Rightarrow \triangle KCQ \cong \triangle DAQ$. So $Q$ is the spiral similarity of $KD \mapsto CA$ and send $B \mapsto X$, so $A$ is circumcenter of $\triangle BXC$ Now only angle-chasing; since $\triangle QBX \sim \triangle QDA$ we get $\angle QBX=\angle QCA$ if $N \equiv AC \cap BX ( BQNC$ is cyclical) so $\angle BNC=\angle BXC + \angle ACX =\frac{\angle BAC}{2}+\frac{\angle BDC}{2}= \angle BQC$ analogously $\angle BPC$ is equal. [asy][asy] import graph; size(12cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -15.993367309653488, xmax = 16.923538317236552, ymin = -1.7033299354902829, ymax = 22.94243471893839; /* image dimensions */ pen zzttqq = rgb(0.6,0.2,0.); draw((-10.05312859066007,9.126739919767129)--(-14.280461815106623,9.301538961734716)--(-0.8757367961679552,0.6082607488077892)--cycle, zzttqq); draw((-7.011038355607665,21.273125988778688)--(-10.05312859066007,9.126739919767129)--(-8.169315038406285,5.338317108632208)--cycle, blue); /* draw figures */ draw((-10.05312859066007,9.126739919767129)--(-0.8757367961679552,0.6082607488077892)); draw((-10.05312859066007,9.126739919767129)--(-7.011038355607665,21.273125988778688)); draw((-7.011038355607665,21.273125988778688)--(-13.904960240604732,0.8488164128106579), red); draw((-7.011038355607665,21.273125988778688)--(-0.8757367961679552,0.6082607488077892), red); draw((-14.280461815106623,9.301538961734716)--(-0.8757367961679552,0.6082607488077892)); draw((-8.169315038406285,5.338317108632208)--(-13.904960240604732,0.8488164128106579)); draw((-14.280461815106623,9.301538961734716)--(-13.904960240604732,0.8488164128106579)); draw((-13.904960240604732,0.8488164128106579)--(-0.8757367961679552,0.6082607488077892)); draw((-10.05312859066007,9.126739919767129)--(-14.280461815106623,9.301538961734716)); draw((-10.05312859066007,9.126739919767129)--(-13.904960240604732,0.8488164128106579)); draw((-10.05312859066007,9.126739919767129)--(-14.280461815106623,9.301538961734716), zzttqq); draw((-14.280461815106623,9.301538961734716)--(-0.8757367961679552,0.6082607488077892), zzttqq); draw((-0.8757367961679552,0.6082607488077892)--(-10.05312859066007,9.126739919767129), zzttqq); draw((-7.011038355607665,21.273125988778688)--(-10.05312859066007,9.126739919767129), blue); draw((-10.05312859066007,9.126739919767129)--(-8.169315038406285,5.338317108632208), blue); draw((-8.169315038406285,5.338317108632208)--(-7.011038355607665,21.273125988778688), blue); draw((14.5256915372546,20.352340382183012)--(-13.904960240604732,0.8488164128106579)); draw((-7.011038355607665,21.273125988778688)--(14.5256915372546,20.352340382183012), red); draw((14.5256915372546,20.352340382183012)--(-0.8757367961679552,0.6082607488077892)); /* dots and labels */ dot((-13.904960240604732,0.8488164128106579),dotstyle); label("$B$", (-14.484442943055813,-0.3061777441961404), NE * labelscalefactor); dot((-0.8757367961679552,0.6082607488077892),dotstyle); label("$C$", (-0.12171841655200773,-0.6973803577585003), NE * labelscalefactor); dot((-8.169315038406285,5.338317108632208),dotstyle); label("$D$", (-8.113428950754514,4.10882318029335), NE * labelscalefactor); dot((-7.011038355607665,21.273125988778688),dotstyle); label("$A$", (-6.772162847112135,21.601168615296015), NE * labelscalefactor); dot((-14.280461815106623,9.301538961734716),dotstyle); label("$K$", (-14.931531644269938,9.976862383728749), NE * labelscalefactor); dot((-10.05312859066007,9.126739919767129),dotstyle); label("$Q$", (-10.79596115803927,9.641545857818153), NE * labelscalefactor); dot((14.5256915372546,20.352340382183012),dotstyle); label("$X$", (15.079297424728283,19.70104163513598), NE * labelscalefactor); dot((-3.1397249632262088,8.233804485148607),linewidth(2.pt) + dotstyle); label("$N$", (-2.748364536184999,7.797304965309886), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy]

My solution: The inversion $I_A$ gives us the equivalent problem: $\triangle{ABC}$ has $AB = AC$. $D$ is the point such that $DB+DC = AB = AC$. $P, Q$ are on $DB, DC$ such that $BP = BA = CA = CQ$. Prove that: $B, C, P, Q$ are concyclic. Proof. We have: $DP = BP-BD = BA-BD = DC$ and analogously, we have: $DQ = DB$ $\Rightarrow BCPQ$ is the isoceles trapezium, and the conclusion follows. Q.E.D

It is well known that $P$ is the midpoint of the arc $ADB$ of $\odot ABD$ Simialrly, $Q$ is the midpoint of arc $ADC$ of $\odot ACD$ Now invert with centre $D$and radius $DA$. Let the image of $X$ be denoted by $X'$. Now since $A=A'$, $P'$ lies on line $AB'$ Also, $Q'$ lies on line $AC'$ Now we have to prove that $B'C'P'Q'$ are concyclic. That is we have to show that, $AB' \cdot AP' = AC' \cdot AQ'$ Now we know that $AP' = \frac{AD \cdot AP}{PD}$, $AQ' = \frac{AD \cdot AQ}{QD}$ $AC' = \frac{AD \cdot AC}{CD}$ $AB' = \frac{AD \cdot AB}{BD}$ Thus we are left to prove that $$\frac{AP}{PD}\cdot \frac{AB}{BD} = \frac{AQ}{QD} \cdot \frac{AC}{CD}$$That is $$AP \cdot CD \cdot DQ = AQ \cdot PD \cdot BD$$ Applying Ptolemy's theorem to cyclic quadrilateral $APDB$, $$AP \cdot BD + PD \cdot AB = AD \cdot BP$$Using, $DA = DB+DC$ and $PA=PB$, $$PD \cdot AB = CD \cdot AP$$ Similarly, $$QD \cdot AC = BD \cdot AQ$$ Since $AB=AC$, $$\frac{PD}{QD}=\frac{CD \cdot AP}{BD \cdot AQ}$$That is $$AP \cdot CD \cdot DQ = BD \cdot AQ \cdot PD$$ QED

Lemma(Extention of Archemedes Midpoint Theorem): If $M$ is the midpoint of arc $BAC$, then \[\frac {MP}{MA}=\frac {|AB-AC|}{BC}\] Main Proof: We quickly see that $ABDP, ACDQ$ are cyclic quadrilaterals. By radical axis theorem, $B,C,Q,P$ are concyclic $\iff QC,BP,AD$ are concurrent. By letting $BP\cap AD=E$, $CQ\cap AD=F$, it suffices to show \[\frac {DE}{AE}=\frac {DF}{AF}\]. Applying the lemma on $\triangle ADB$ and midpoint $P$, we have \[\frac {PD}{AP}=\frac {PD}{BP}=\frac {AD-DB}{AB}=\frac {DC}{AB}\]Hence we have \[\frac {DE}{AE}=\frac {BD}{AB}\cdot \frac {\sin \angle DBP}{\sin \angle ABP}=\frac {BD}{AB}\cdot \frac {PD}{AP}=\frac {BD\cdot DC}{AB^2}\] By symmetry, we can similarly derive $\frac {DF}{AF}=\frac {BD\cdot DC}{AC^2}=\frac {BD\cdot DC}{AB^2}=\frac {DE}{AE}$, and the ratio equality is proven.

Invert around $A$ with radius $AB$. Denote the inverse by a $^*$. $D^*A$ = $AB.DB/AD$ etc imply $D^*B + D^*C = AB$. We recognise $P$ as the midpoint of arc $ADB$, so it is on the intersection of it and the perpendicular bisector of $BC$. Thus its inverse is the intersection of $BD^*$ and circle with $B$ as center and $AB$ as radius. Now we compute $D^*Q^* . D^*C = (AB-D^*C)D^*C = AB.D^*C - D^*C^2$. By a similar computation for the other product get $B, C, P^*, Q^*$ are concyclic. Inverting back, we have the conclusion.

先证明A、P、D、B四点共圆 过P做AD、BD垂线，垂足为X、Y 显然PX=PY、AP=BP、∠AXP=∠AYP ∴AXP、BYP全等 ∠PAX=∠PBY ∴A、P、D、B共圆 同理 A、Q、D、C共圆 下证BP、CQ、AD共点 设BP交AD于T1，CQ交AD于T2 只需证AT1/T1D=AT2/T2D 延长DB至K使得KB=DC，连接AK 显然APB、ADK相似 立得APD、ABK相似 ∴AT1/T1D=（AB/BD）*（sin∠ABT1/sin∠DBT1）=（AB/BD）*（sin∠ADP/sin∠PAD）=（AB/BD）*（AP/PD）=（AB/BD）*（AB/BK）=（AB*AC）/（BD*CD） 同理AT2/T2D=（AB*AC）/（BD*CD） ∴T1=T2（称这点为T） ∴BP、CQ、AD交于同一点T ∴BT*TP=AT*TD=CT*TQ 即B、P、C、Q四点共圆 Q.E.D.

I wonder if a solution along the lines of "...cevians $\overline{BP}, \overline{CQ}, \overline{DA}$ concur hence by radical axis..." exists. dibyo_99 wrote: $\triangle{ABC}$ is isosceles with $AB = AC >BC$. Let $D$ be a point in its interior such that $DA = DB+DC$. Suppose that the perpendicular bisector of $AB$ meets the external angle bisector of $\angle{ADB}$ at $P$, and let $Q$ be the intersection of the perpendicular bisector of $AC$ and the external angle bisector of $\angle{ADC}$. Prove that $B,C,P,Q$ are concyclic. Invert at $D$. Then $D$ lies inside $\triangle A'B'C'$ with $\frac{A'B'}{A'C'}=\frac{DB'}{DC'}$ and $\frac{1}{DA'}=\frac{1}{DB'}+\frac{1}{DC'}$; external bisectors of angles $A'DB', A'DC'$ meet lines $A'B'$ and $A'C'$ again at $P',Q'$. It suffices to show $B',C',P',Q'$ are concyclic. Equivalently, we want $A'B' \cdot A'P'=A'C' \cdot A'Q'$. By external bisector theorem, $\frac{A'P'}{A'B'}=\frac{A'D}{B'D-A'D}=\frac{DC'}{DB'}$ and similarly $\frac{A'Q'}{A'C'}=\frac{DB'}{DC'}$. The claim now follows.

This problem is trivial with inversion but very nice to solve without it . Here is my synthetic solution. Clearly quadrilaterals $APDB$ and $AQDC$ are cyclic. Extend $DB$ beyond $B$ and $DC$ beyond $C$ to $X,Y$ respectively so that $DA=DX=DY$. Then $\triangle ADX\stackrel{+}{\sim}\triangle APB$ $\implies \triangle APD\stackrel{+}{\sim}\triangle ABX$. Similarly, we get $\triangle AQD\stackrel{+}{\sim}\triangle ACY$. Now construct point $U$ such that $\triangle DAU\stackrel{+}{\sim} \triangle DPC$. Hence $\triangle DCU\stackrel{+}{\sim} \triangle DPA \sim\triangle XBA$. But $BX=DC$ so $\triangle DCU\cong\triangle XBA$ or $CU=CA$. Similarly if we construct point $V$ such that $\triangle DAV\stackrel{+}{\sim} \triangle DQB$, we also get $BV=BA$. Moreover, $$\angle BPC = \angle BPD + \angle DPC = \angle BAD + \angle DAU = \angle BAU$$Similarly $\angle BQC = \angle CAV$. Hence it suffices to show that $\triangle ABV\cong\triangle ACU$. But $$\angle ACU = \angle DCU - \angle ACD = \angle DPA - \angle ACD = 180^{\circ} - \angle ABD - \angle ACD$$which is symmetric w.r.t. $B,C$ so we are done.

First intersect perpendicular from A , B to AD , BD in K , perpendicular from A , C to AD , CD in J , perpendicular from B , C to BD , CD in I . Bisector of K , J meet IJ , IK in M ,L . It's easy to find that D is on the line LM (because AD=DB+DC) We know that P , K , M and Q , J , L are collinear ( because ADBK , ADCJ are cyclic ) It's easy to see that DPMC , DDQLB are cyclic . Now we have BPC=BAD+DMC , BQC=DAC+DLB . we should just prove that BAD+DMC=DAC+DLB and it's equal to prove that JMD=KDL . If LM intersect KJ at X , we know that XJ.KI=XK.IJ and we want to say that DJ.XK=DK.XJ so it's enough to prove IJ.DK=DJ.IK . Because AC=AB we can easily prove IJ.DK=DJ.IK so the problem is solved