This seems very false, just consider when $BC||QR$ like so: [asy][asy] /* File unicodetex not found. */ /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(13.18cm); 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 = 20.18, xmax = 33.36, ymin = 21.28, ymax = 31.41; /* image dimensions */ Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 2, Size = 2, NoZero), Arrows(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 2, Size = 2, NoZero), Arrows(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((24.13,25.49)--(28.38,25.48)); draw((21.82,26.77)--(26.25,25.49)); draw((26.25,25.49)--(30.7,26.76)); draw((26.25,25.49)--(26.26,26.76)); draw((24.13,25.49)--(26.26,26.76)); draw((26.26,26.76)--(28.38,25.48)); draw((21.82,26.77)--(30.7,26.76)); /* dots and labels */ dot((26.26,26.76),dotstyle); label("$A$", (26.35,26.9), NE * labelscalefactor); dot((24.13,25.49),dotstyle); label("$B$", (24.23,25.63), NE * labelscalefactor); dot((28.38,25.48),dotstyle); label("$C$", (28.46,25.63), NE * labelscalefactor); dot((26.25,25.49),dotstyle); label("$P$", (26.35,25.63), NE * labelscalefactor); dot((21.82,26.77),dotstyle); label("$Q$", (21.91,26.9), NE * labelscalefactor); dot((30.7,26.76),dotstyle); label("$R$", (30.79,26.9), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] I think this problem need the condition $BC\not{||}QR$ to be true. Edit: Ah, right. Thanks.

Or perhaps by "bisect", the problem meant an internal angle bisector, rather than an external one.

Let $(O), (D)$ be circumcircles of $\triangle ABC, \triangle PQR,$ resp. Let perpendicular bisectors of $|BC|, |QR|$ meet at $X$ and let $Y \equiv BC \cap XA$ $\Longrightarrow$ $X \in (O) \cap (D)$ and $Y \in (D)$. Let $C' \in AB$ be reflection of $C$ in $XA$ $\Longrightarrow$ $|AC'| = |AC|$ and perpendicular bisectors $XP, XA$ of $|BC|, |CC'|$ meet at circumcenter $X$ of $\triangle BCC'$. $\triangle AXB \sim \triangle ACY$ are similar, having equal angles $\Longrightarrow$ $\overline{AB} \cdot \overline{AC'} = \overline{AB} \cdot \overline{AC} = \overline{AX} \cdot \overline{AY} = \overline{AQ} \cdot \overline{AR}$ $\Longrightarrow$ $BQC'R$ is cyclic. Isosceles trapezoid $QCC'R$ is also cyclic $\Longrightarrow$ $BQCC'R$ is cyclic with circumcircle $(X)$ of $\triangle BCC'.$ Let $PR$ cut $(X)$ again at $S$. Since $P$ is midpoint of $|BC|$ and $BC$ bisects $\angle RPQ$ $\Longrightarrow$ $S$ is reflection of $Q$ in perpendicular bisector $XP$ of $|BC|$ $\Longrightarrow$ $|PS| = |PQ|$ and $BSQC$ is isosceles trapezoid $\Longrightarrow$ $BSC'R$ is also isosceles trapezoid and $|AB| + |AC| = |AB| + |AC'| = |BC'| = |SR| = |PS| + |PR| = |PQ| + |PR|$.

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