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Concyclic with B,C

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Source: China Team Selection Test 3 Day 1 P1

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487 posts

#1 h Mar 25, 2015, 2:13 PM • 6 Y

Y by tenplusten, Davi-8191, nguyendangkhoa17112003, Adventure10, Mango247, Rounak_iitr

$\triangle{ABC}$ 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.

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#2 h Mar 25, 2015, 5:01 PM • 6 Y

Y by langkhach11112, dagezjm, mathisreal, SMSGodslayer, Adventure10, Mango247

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

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andria

#3 h Mar 25, 2015, 7:25 PM • 4 Y

Y by Virgil Nicula, Davi-8191, Adventure10, Mango247

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 $A'Q'=A'B'$ .

This post has been edited 1 time. Last edited by andria, Mar 27, 2015, 6:14 PM

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#4 h Apr 3, 2015, 8:02 PM • 3 Y

Y by tenplusten, Adventure10, Mango247

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.

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#5 h Apr 4, 2015, 12:45 AM • 4 Y

Y by Devesh14, Adventure10, Mango247, Rounak_iitr

We begin with a well-known lemma that a few above have cited:
Lemma
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. $\square$

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yugrey

#6 h Apr 4, 2015, 6:30 PM • 5 Y

Y by TripteshBiswas, tenplusten, Adventure10, Mango247, Infinityfun

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.

This post has been edited 2 times. Last edited by yugrey, Apr 4, 2015, 6:31 PM
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#8 h Apr 18, 2015, 6:36 PM • 2 Y

Y by Adventure10, Mango247

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.

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#9 h May 24, 2015, 11:58 PM • 7 Y

Y by tranquanghuy7198, langkhach11112, Uagu, GGPiku, k12byda5h, Adventure10, Mango247

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.

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tranquanghuy7198

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tranquanghuy7198

#10 h May 25, 2015, 4:28 AM • 5 Y

Y by langkhach11112, thedragon01, Adventure10, Mango247, Infinityfun

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

Attachments:

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#11 h Jan 8, 2016, 1:54 PM • 4 Y

Y by langkhach11112, anantmudgal09, Adventure10, Mango247

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

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XmL

#12 h May 8, 2016, 7:21 PM • 2 Y

Y by parola, Adventure10

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.

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#13 h Feb 21, 2017, 6:47 AM • 2 Y

Y by Adventure10, Mango247

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.

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#14 h Feb 22, 2017, 3:30 PM • 2 Y

Y by Adventure10, Mango247

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.

This post has been edited 1 time. Last edited by dagezjm, Feb 22, 2017, 3:33 PM
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#15 h Mar 25, 2018, 10:29 PM • 2 Y

Y by Adventure10, Mango247

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 post has been edited 1 time. Last edited by anantmudgal09, Mar 25, 2018, 10:29 PM

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#16 h Oct 16, 2018, 9:43 AM • 2 Y

Y by Adventure10, Mango247

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.

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#17 h Apr 29, 2020, 8:36 AM

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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

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#18 h Apr 20, 2024, 8:21 PM

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Long solution but with some nice other properties of the config

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