$P,Q$ lie on perpendicular bisectors of $DE \Rightarrow PQ \parallel BC$ [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(17.54480626755017cm); 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 = -15.625438176669318, xmax = 32.91936809088085, ymin = -15.951634658044359, ymax = 9.413601792180287; /* image dimensions */ /* draw figures */ draw((xmin, -1.892916496079884*xmin + 10.322773993332826)--(xmax, -1.892916496079884*xmax + 10.322773993332826), linewidth(0.4)); /* line */ draw((3.5852293375024153,-2.3745965978382966)--(10.059642312394205,-0.7647064419390989), linewidth(0.4)); draw((6.345273995976908,-1.6882998257985868)--(6.320080999135746,-7.129987143489912), linewidth(0.4)); draw((6.345273995976908,-1.6882998257985868)--(3.4805879745768706,-4.395939366482308), linewidth(0.4)); draw((6.345273995976908,-1.6882998257985868)--(16.117981007943325,-4.454445815710853), linewidth(0.4)); draw((3.339765411491172,-7.116189386139705)--(0.6159019531768349,-7.103578907166027), linewidth(0.4)); draw((3.4805879745768706,-4.395939366482308)--(6.320080999135746,-7.129987143489912), linewidth(0.4)); draw((6.320080999135746,-7.129987143489912)--(16.117981007943325,-4.454445815710853), linewidth(0.4)); draw(circle((12.001374761685923,-2.488569744074393), 9.820302443764644), linewidth(0.4)); draw((0.6159019531768349,-7.103578907166027)--(-15.139727127088342,-7.030636179942577), linewidth(0.4)); draw((12.910324337163027,-2.50087112956549)--(-15.139727127088342,-7.030636179942577), linewidth(0.4)); draw((10.059642312394205,-0.7647064419390989)--(3.339765411491172,-7.116189386139705), linewidth(0.4)); draw((3.5852293375024153,-2.3745965978382966)--(20.61976541149118,-7.1961893861397055), linewidth(0.4)); draw((3.4805879745768706,-4.395939366482308)--(16.117981007943325,-4.454445815710853), linewidth(0.4)); draw((3.86391356954328,3.008708258117443)--(3.339765411491172,-7.116189386139705), linewidth(0.4)); draw((3.339765411491172,-7.116189386139705)--(20.61976541149118,-7.1961893861397055), linewidth(0.4)); draw((20.61976541149118,-7.1961893861397055)--(3.86391356954328,3.008708258117443), linewidth(0.4)); draw((3.5852293375024153,-2.3745965978382966)--(-15.139727127088342,-7.030636179942577), linewidth(0.4)); /* dots and labels */ dot((3.339765411491172,-7.116189386139705),dotstyle); label("$A$", (3.441627318227462,-6.835104097135953), NE * labelscalefactor); dot((3.86391356954328,3.008708258117443),dotstyle); label("$B$", (3.988043976452257,3.2879834657654925), NE * labelscalefactor); dot((20.61976541149118,-7.1961893861397055),dotstyle); label("$C$", (20.72564898102228,-6.9213804115925), NE * labelscalefactor); dot((10.059642312394205,-0.7647064419390989),linewidth(4.pt) + dotstyle); label("$A_1$", (9.509728101671234,-0.07679279803981758), NE * labelscalefactor); dot((9.227147533637993,-7.1434457848533475),linewidth(4.pt) + dotstyle); label("$B_1$", (9.33717547275814,-6.9213804115925), NE * labelscalefactor); dot((3.5852293375024153,-2.3745965978382966),linewidth(4.pt) + dotstyle); label("$C_1$", (2.8952106600026677,-1.7160427727141994), NE * labelscalefactor); dot((6.345273995976908,-1.6882998257985868),linewidth(4.pt) + dotstyle); label("$D$", (6.288745695293497,-1.4859726008300755), NE * labelscalefactor); dot((6.320080999135746,-7.129987143489912),linewidth(4.pt) + dotstyle); label("$E$", (6.4325395527210745,-6.892621640106984), NE * labelscalefactor); dot((3.4805879745768706,-4.395939366482308),linewidth(4.pt) + dotstyle); label("$P$", (2.923969431488183,-4.8219900931498705), NE * labelscalefactor); dot((16.117981007943325,-4.454445815710853),linewidth(4.pt) + dotstyle); label("$Q$", (16.06672800036877,-3.901709405613375), NE * labelscalefactor); dot((7.256852924643081,-3.413842617349615),linewidth(4.pt) + dotstyle); label("$I$", (7.29530269728654,-3.153981346989973), NE * labelscalefactor); dot((-15.139727127088342,-7.030636179942577),linewidth(4.pt) + dotstyle); label("$R$", (-15.021503975473491,-6.806345325650438), NE * labelscalefactor); dot((12.910324337163027,-2.50087112956549),linewidth(4.pt) + dotstyle); label("$T$", (13.018298222904125,-2.262459430938993), NE * labelscalefactor); dot((9.781838946264951,-8.193430310048775),linewidth(4.pt) + dotstyle); label("$O$", (9.883592130982935,-7.956696185071056), NE * labelscalefactor); dot((7.786210764807451,-4.415872805325967),linewidth(4.pt) + dotstyle); label("$K$", (7.899236898482366,-4.18929712046853), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Claim: $D$ is incenter of $\triangle APQ$ Let $K=DB_1 \cap PQ$. $K$ is midpoint of $DB_1$. If $R=A_1C_1 \cap AC$ and $T=IR \cap BC$ then by ceva's harmoinc property $(B,T;A_1,C)\stackrel{R}{=}(B,I;D,B_1)=-1$.Now from harmonic property $$BI.BK=BD.BB_1 \Rightarrow \frac{BK}{BD}=\frac{BB_1}{BI} \Rightarrow \frac{DK}{BD}=\frac{IB_1}{BI}$$ From $\frac{BD}{DK}=\frac{BI}{IB_1}$ and $\triangle PBQ \sim \triangle ABC$ we prove our claim. Now by incenter-excenter lemma circumcenter of $\triangle DPQ$ lie on angle bisector of $\measuredangle{PAQ}$ which is $BD$ hence $O$ lie on $BD$ As $DE$ is altitude from $D$ to $PQ$ and it well know $\measuredangle{PDE}=\measuredangle{ODQ}$ and hence we get $\measuredangle{PDB_1}=\measuredangle{EDQ}$.

Neat! starchan wrote: Let $AA_1, BB_1, $ and $CC_1$ be the bisectors of a triangle $ABC$. The segments $BB_1$ and $A_1C_1$ meet at point $D$. Let $E$ be the projection of $D$ to $AC$. Points $P$ and $Q$ on sides $AB$ and $BC$ respectively are such that $EP = PD, EQ = QD$. Prove that $\angle PDB_1 = \angle EDQ$. Let $L$ be the foot of the $B$-angle bisector on line $AC$ and $K$ be the midpoint of $DL$ and $I$ be the incentre of $ABC$. Notice that $(BI, DL)=-1$ and so $KD^2=KI \cdot KA$ and in fact $BI \cdot BK = BD \cdot BL$ or $\frac{BI}{BL} = \frac{BD}{BK}$. Since $K$ lies on $PQ$ and $PQ \parallel AC$, we conclude that $\triangle BPQ \sim \triangle BAC$ under a homothety at $B$ which maps $D$ to $I$ and $K$ to $L$. In particular, since $K$ is the foot of $B$-angle bisector in $BPQ$, we conclude that $D$ is the incentre of $\triangle BPQ$. Now by Fact 5, lines $DB$ and $DE$ are isogonal in angle $PDQ$, from which the conclusion follows as $D, B, B_1$ are collinear.

Let $BB_1$ meet $PQ$ at $F$. Claim $: D$ is incenter of $APQ$. Proof $:$ Note that $(BI,DB_1) = -1$ and since $PQ$ is perpendicular bisector of $DE$ and $AC \parallel PQ$ we have that $F$ is midpoint of $DB_1$ so $BD.BB_1 = BI.BF$ and since $BAC$ and $BPQ$ are similar we have that $D$ is incenter. Now $\angle PDE = 90 - \angle DPQ = \angle DQB + \angle DBQ = \angle QDB_1$ so $\angle PDB_1 = \angle QDE$.