Should it be SQ = RB instead of RB'?

Vo Duc Dien wrote: Should it be SQ = RB instead of RB'? No, the problem statement is true.

I guess you have to be able to solve the problem first before being able to graph it. It's the reverse process to say the least. I have not been able to draw the graph less alone solve it. If you're not able to graph it, you will not solve it. Dien

Can someone solve for the last time with a graph please ?

crazyfehmy wrote: Let $P$ be an interior point of the triangle $ABC$ different from the centroid and satisfying $\angle CAP = \angle BCP. \: BP \cap CA = \{B'\} \: , \: CP \cap AB = \{C'\}$ and $Q$ is the second point of intersection of $AP$ and the circumcircle of $ABC. \: B'Q$ intersects $CC'$ at $R$ and $B'Q$ intersects the line through $P$ parallel to $AC$ at $S.$ Let $T$ be the point of intersection of lines $B'C'$ and $QB$ and $T$ be on the other side of $AB$ with respect to $C.$ Prove that \[\angle BAT = \angle BB'Q \: \Longleftrightarrow \: |SQ|=|RB'| \] But $S,Q,R,B'$ are collinear and $R,B'$ lie between $S,Q!$ Please refer to my diagram for the counterexample... it seems Geogebra does not agree to the problem statement. Can you please correct the problem? [asy][asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ffcccc = rgb(1,0.8,0.8); pen qqwuqq = rgb(0,0.39,0); draw((0,5.92)--(-2,0)--(5,0)--cycle); draw((0,5.92)--(-2,0),blue); draw((-2,0)--(5,0),blue); draw((5,0)--(0,5.92),blue); pair parametricplot0_cus(real t){ return (0.94*cos(t)+5,0.94*sin(t)+0); } draw(graph(parametricplot0_cus,2.7509491252437344,3.141592653589793)--(5,0)--cycle,qqwuqq); pair parametricplot1_cus(real t){ return (0.94*cos(t)+0,0.94*sin(t)+5.92); } draw(graph(parametricplot1_cus,-1.2578093541863007,-0.8695918785813073)--(0,5.92)--cycle,qqwuqq); draw(circle((5.01,5.08),5.08),red); draw((-2,(+2.93+1.47*-2)/3.44)--(7.66,(+2.93+1.47*7.66)/3.44),blue); draw((-12.21,(-7.33+1.47*-12.21)/-3.56)--(5,(-7.33+1.47*5)/-3.56),blue); draw(circle((1.5,2.12),4.09),red); draw((0,(+8.54-4.46*0)/1.44)--(7.66,(+8.54-4.46*7.66)/1.44),blue); draw((2.51,(-11.32+4.04*2.51)/0.64)--(7.66,(-11.32+4.04*7.66)/0.64),blue); draw((-12.21,(+5.61-5.92*-12.21)/-2)--(7.66,(+5.61-5.92*7.66)/-2),blue); draw((-12.21,(-10.48+0.34*-12.21)/-4.29)--(3.15,(-10.48+0.34*3.15)/-4.29),blue); draw((-12.21,(+3.69+1.85*-12.21)/-4.51)--(2.51,(+3.69+1.85*2.51)/-4.51),blue); draw((-9.87,3.22)--(0,5.92),blue); pair parametricplot2_cus(real t){ return (0.94*cos(t)+3.15,0.94*sin(t)+2.19); } draw(graph(parametricplot2_cus,-2.739196770479163,-1.7269307767856577)--(3.15,2.19)--cycle,red+linewidth(1.2pt)); pair parametricplot3_cus(real t){ return (0.94*cos(t)+0,0.94*sin(t)+5.92); } draw(graph(parametricplot3_cus,-2.874409099783693,-1.8965074146760865)--(0,5.92)--cycle,red+linewidth(1.2pt)); dot((0,5.92),ds); label("$A$", (0.13,6.12),NE*lsf); dot((-2,0),ds); label("$B$", (-2.43,-0.56),NE*lsf); dot((5,0),ds); label("$C$", (5.13,-0.56),NE*lsf); dot((1.44,1.47),ds); label("$P$", (1.01,0.97),NE*lsf); dot((3.15,2.19),ds); label("$B'$", (3.28,2.37),E*lsf); dot((-1.15,2.53),ds); label("$C'$", (-1.52,2.84),NE*lsf); dot((2.51,-1.85),ds); label("$Q$", (2.29,-2.44),E*lsf); dot((2.94,0.85),ds); label("$R$", (3.07,1.03),E*lsf); dot((4.43,10.3),ds); label("$S$", (3.94,10.24),E*lsf); dot((-9.87,3.22),ds); label("$T$", (-9.74,3.4),NE*lsf); clip((-12.21,-4.09)--(-12.21,11.83)--(7.66,11.83)--(7.66,-4.09)--cycle);[/asy][/asy]

I think you misunderstood the question. $S$ is the intersection point of $B'Q$ and the line through $P$ and parallel to $AC$ not $AB.$

Parallels to $AB, AC$ through $P$ cut $QB$ at $X, Y$ and $AC$ itself cuts $QB$ at $Z.$ $\angle CBQ = \angle CAQ = \angle CAP = \angle BCP$ $\Longrightarrow$ $PC \parallel QB$ $\Longrightarrow$ $\frac{\overline{QX}}{\overline{QY}} = \frac{\overline{QB}}{\overline{QZ}}$ and $\frac{\overline{BT}}{\overline{BZ}} = \frac{\overline{PC'}}{\overline{PC}} = \frac{\overline{QB}}{\overline{QZ}} = \frac{\overline{RP}}{\overline{RC}}.$ It follows that $PX, RQ, CY$ are concurrent (or parallel). Equality of these ratios holds for any $P,$ such that $\angle CAP = \angle BCP.$ Then: $\overline{QS} = \overline{RB'}$ $\Longleftrightarrow$ $\triangle QYS \cong \triangle RCB'$ $\Longleftrightarrow$ $CY \parallel B'RSQ \parallel PX \parallel AB$ $\Longleftrightarrow$ $\frac{\overline{BT}}{\overline{BZ}} = \frac{\overline{QB}}{\overline{QZ}} = \frac{\overline{B'A}}{\overline{B'Z}}$ $\Longleftrightarrow$ $BB' \parallel AT$ $\color{red}{\Longrightarrow}$ $\angle BAT = \angle QB'B.$ Conversely, $\color{red}{\angle BAT = \angle QB'B}$ $\color{red}{\not\Longrightarrow}$ $\color{red}{\overline{QS} = \overline{RB'}}$. Equality of these angles combined with $\frac{\overline{BT}}{\overline{BZ}} = \frac{\overline{QB}}{\overline{QZ}}$ implies just central similarity of circumcircles $\odot(BAT) \sim \odot(QB'B)$ with center $Z.$ However, the points $A, B'$ collinear with this similarity center are not necessarily corresponding points on these circumcircles, because the line $ZB'A$ cuts each of these circles in 2 points, so that $\triangle BAT, \triangle QB'B$ are not necessarily centrally similar $\Longrightarrow$ lines $CY, B'RSQ$ are not necessarily parallel $\Longrightarrow$ segments $\overline{QS}, \overline{RB'}$ are not necessarily equal.

Attachments:

Actually, I have realized that I wrote the problem with some mistake. I should have said that $P$ is not on the median of $BC$ instead of saying centroid. Now, I think your counter example does not work. Sorry for this typo.

crazyfehmy wrote: Actually, I have realized that I wrote the problem with some mistake. I should have said that $P$ is not on the median of $BC$ instead of saying centroid... In that case, assume $\angle BAT = \angle QB'B$ and $BA \not\parallel QB', TA \not\parallel BB'$. Since $Z$ is external similarity center of the circles $\odot(BAT), \odot(QB'B)$, inversion with center $Z$ and power $ZB^2$ takes one circle to the other $\Longrightarrow$ $\overline{ZQ} \cdot \overline{ZT} = ZB^2 = \overline{ZB'} \cdot \overline{ZA}$ $\Longrightarrow$ $ATQB'$ is cyclic $\Longrightarrow$ $\angle BCC' = \angle BCP = \angle CAP = \angle B'AQ = \angle B'TQ = \angle B'C'C$ $\Longrightarrow$ $B'C' \parallel BC$ $\Longrightarrow$ $P$ is on A-median of $\triangle ABC,$ which contradicts the additional problem condition.

Here is a solution by Burak Varıcı. $PS \parallel AC$ and $\angle CAP=\angle BCP$ gives $\angle QPC=\angle ACB=\angle AQB=\angle PQB$ concludes in $BQ \parallel PC$ Let $AP \cap BC= A'$, $B'C' \cap AP= U$ and $(B'PC') \cap AP =D$. We will first prove $D \neq A$. Assume contrary. Then we have $A \in (B'C'P)$ Thus $\angle PAB' = \angle B'C'P = \angle PCA' \Rightarrow B'C' \parallel BC \Rightarrow \angle C'AP = \angle C'B'P = \angle PBC$ results in $A'B^{2}=A'P \cdot A'A$. Similarly $A'C^{2} = A'P \cdot A'A$. So $A'B=A'C$. However $AP$ is not median. Contradiction $D \neq A$. We will prove when $D \in [AU]$. $A \in [DU]$ can be proven in similar way. We will first prove $SQ=RB' \Leftrightarrow AB \parallel B'Q$, then we will prove $AB \parallel B'Q \Leftrightarrow \angle BAT = \angle BB'Q$ Since $PS \parallel AB'$, we have $\frac{PQ}{PA}=\frac{SQ}{SB'}$. If $SQ=RB'$, then because of $BQ \parallel PC$ $\frac{PQ}{PA}=\frac{RB'}{RQ}=\frac{PB'}{PB} \Rightarrow AB \parallel B'Q$ If $AB \parallel B'Q$, we will use $BQ \parallel PC$ $\frac{RB'}{RQ}=\frac{PB'}{PB}=\frac{PQ}{PA}=\frac{SQ}{SB'} \Rightarrow SQ=RB'$ Hence $SQ=RB' \Leftrightarrow AB \parallel B'Q$ Since $PC' \parallel QT$, we have $\frac{UD}{UB'}=\frac{UC'}{UP}=\frac{UT}{UQ}$. Therefore $\triangle TDU \sim \triangle QB'U$ and $\triangle TDC' \sim \triangle QB'P$ If $\angle BAT= \angle BB'Q$, then $\angle C'AT=\angle BAT=\angle BB'Q=\angle PB'Q=\angle C'DT \Rightarrow T, A, D, C' $ are cyclic. So $\angle PAB=\angle DAC'=\angle DTC'=\angle B'QP \Rightarrow AB \parallel B'Q$ If $AB \parallel B'Q$, we will see $\angle DTC' = \angle B'QP= \angle DAC' \Rightarrow T, A, D, C'$ are cyclic.Thus $\angle BAT{\rm =}\angle C'AT=\angle C'DT=\angle PB'Q=\angle BB'Q$ Hence $AB \parallel B'Q \Leftrightarrow \angle BAT= \angle BB'Q$ In conclusion $SQ=RB' \Leftrightarrow AB \parallel B'Q \Leftrightarrow \angle BAT=\angle BB'Q$