[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(300); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.000000000000000, xmax = 12.50000000000001, ymin = -0.7500000000000009, ymax = 9.000000000000000; /* image dimensions */ /* draw figures */ draw((5.920000000000006,3.625721223959009)--(5.244697032519201,-0.004640279880253464)); draw((5.244697032519201,-0.004640279880253464)--(12.00000000000000,0.000000000000000)); draw((8.068866156605040,2.344275560891563)--(5.244697032519201,-0.004640279880253464)); draw((-1.410608263279613,7.997224836574497)--(12.00000000000000,0.000000000000000)); draw((8.169758762373634,1.053937431860808)--(-1.410608263279613,7.997224836574497)); draw((5.920000000000006,3.625721223959009)--(6.928205295519882,1.395565958941424)); draw((6.928205295519882,1.395565958941424)--(12.00000000000000,0.000000000000000)); draw((-1.410608263279613,7.997224836574497)--(5.244697032519201,-0.004640279880253464)); /* dots and labels */ dot((5.920000000000006,3.625721223959009),dotstyle); label("$A$", (5.983583155759467,3.723330902202111), NE * labelscalefactor); dot((5.244697032519201,-0.004640279880253464),dotstyle); label("$B$", (4.819796702577611,-0.5183644600527996), NE * labelscalefactor); dot((12.00000000000000,0.000000000000000),dotstyle); label("$C$", (11.94033223849291,-0.5030514804056699), NE * labelscalefactor); dot((8.068866156605040,2.344275560891563),dotstyle); label("$D$", (8.127400306357623,2.437040611843221), NE * labelscalefactor); dot((6.552900994336857,2.225740990739365),dotstyle); label("$I_a$", (6.596102341644654,2.314536774666184), NE * labelscalefactor); dot((8.169758762373634,1.053937431860808),dotstyle); label("$I_c$", (8.203965204593272,1.166063301131461), NE * labelscalefactor); dot((-1.410608263279613,7.997224836574497),dotstyle); label("$Q$", (-1.351334095215652,8.087530101634059), NE * labelscalefactor); dot((6.928205295519882,1.395565958941424),dotstyle); label("$P$", (6.871735975292989,0.9057426471302569), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let $P$ be the intersection of lines $AI_a$ and $CI_c$. Because $AI_a$, $CI_c$ are angle bisectors of $\triangle ABC$, $P$ lies on $BD$. By Ceva's Theorem, $AI_c$, $CI_a$, and $PD$ are concurrent, so $(QADC)$ is harmonic. Therefore the pencil $B(QADC)$ is also harmonic. But $BD$ bisects $\angle ABC$, so it immediately follows that $\angle DBQ = 90^\circ$.

Denote $PD \cap {I_a}{I_c} = X$, because $\angle {I_a}D{I_c} = 90^\circ $, $D{I_a}$ is the bisector of $\angle PDA$, so $B(QD;AC) = (QD;AC) = P(QX;{I_a}{I_c}) = (QX;{I_a}{I_c}) = D(QX;{I_a}{I_c}) = - 1$, hence $\angle DBQ = 90^\circ $.

Dear Mathlinkers, also by considering the A, C-excenters wrt the triangle ABD, CBD and working with harmonic division and pencil... Sincerely Jean-Louis

Let $U$ and $V$ points of $\overline{AC}$ such that $\overline{BU}$ and $\overline{BV}$ bisects angles ${\angle ABD}$ and $\angle BDC$ respectively. $Q$ and $D$ are harmonic conjugates w/r $\overline{UV}$ for $\overline{BD}$, $\overline{UI_{C}}$ and $\overline{VI_{A}}$ concur. Now, as $\measuredangle ABU=\measuredangle UBD=\measuredangle DBV=\measuredangle VBC$, we know that $Q$ and $D$ are harmonic conjugates w/r $\overline{AC}$ too, so $\overline{QB}\perp\overline{BD}$. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(300); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -2.000000000000000, xmax = 12.50000000000001, ymin = -0.7500000000000009, ymax = 9.000000000000000; /* image dimensions */ /* draw figures */ draw((5.920000000000006,3.625721223959009)--(5.244697032519201,-0.004640279880253464)); draw((5.244697032519201,-0.004640279880253464)--(12.00000000000000,0.000000000000000)); draw((8.068866156605040,2.344275560891563)--(5.244697032519201,-0.004640279880253464)); draw((-1.410608263279613,7.997224836574497)--(12.00000000000000,0.000000000000000)); draw((8.169758762373634,1.053937431860808)--(-1.410608263279613,7.997224836574497)); draw((5.920000000000006,3.625721223959009)--(6.928205295519882,1.395565958941424)); draw((6.928205295519882,1.395565958941424)--(12.00000000000000,0.000000000000000)); /* dots and labels */ dot((5.920000000000006,3.625721223959009),dotstyle); label("$A$", (5.983583155759467,3.723330902202111), NE * labelscalefactor); dot((5.244697032519201,-0.004640279880253464),dotstyle); label("$B$", (4.819796702577611,-0.5183644600527996), NE * labelscalefactor); dot((12.00000000000000,0.000000000000000),dotstyle); label("$C$", (11.94033223849291,-0.5030514804056699), NE * labelscalefactor); dot((8.068866156605040,2.344275560891563),dotstyle); label("$D$", (8.127400306357623,2.437040611843221), NE * labelscalefactor); dot((6.552900994336857,2.225740990739365),dotstyle); label("$I_a$", (6.596102341644654,2.314536774666184), NE * labelscalefactor); dot((8.169758762373634,1.053937431860808),dotstyle); label("$I_c$", (8.203965204593272,1.166063301131461), NE * labelscalefactor); dot((-1.410608263279613,7.997224836574497),dotstyle); label("$Q$", (-1.351334095215652,8.087530101634059), NE * labelscalefactor); dot((6.928205295519882,1.395565958941424),dotstyle); label("$P$", (6.871735975292989,0.9057426471302569), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */[/asy][/asy]

it's easy by Menelous theorem! You see $ \triangle AIC$ and line $I_aI_cQ$. ($I$ is incenter of the triangle $ABC$).

a slightly different approach: let $(\omega_a,U), (\omega_b,V)$ be the $A$-excenters of $ABD, DBC$ resp. consider points $L,K$ the eximilicenters of $((I_a),(\omega_2))$ and $((I_b), (\omega _2))$, well known that $-1=(L,D;I_a,V)=(K,D;I_c,U)$ since $D$ is the insimillicenter of $((I_a),(\omega_2))$ and $((I_b), \omega _2))$. thus : $LB\perp BD, KB\perp BD$ i.e $L,B,K$ are collinear. now it is easy to see that the eximilicenter of $((\omega_1),(\omega_2)$ is that of $((I_a),(I_b))$ and we're done by Monge,

Let $R$ the intersection of $I_aI_b$ and $BD$ . $Q$ and $R$ are resp.exsimilicenter and insmilicenter of $(I_a),(I_b)$ thus $B(Q,R;I_a,_b)=-1$ but $BR$ is bisector of $\widehat{I_aBI_b}$ therefore $\widehat{QBD}=\frac{\pi}{2}$. R HAS

Let $I_aI_c$ $\cap$ $BD$ = $M$, $AI_a$, $CI_c$ and $BD$ are concurrent in $I$ (incenter of $ABC$). It´s easy to see: $\angle$ $I_aDI_c$ = $90^\circ$ , $DI_a$ and $DI_c$ are bisectriz. ($I_a$,$M$,$I_c$,$Q$)=$-1$. By theorem ($A$,$D$,$C$,$Q$)=$-1$. $BD$ is bisectriz $\longrightarrow$ $\angle$ $DBQ$ = $90^\circ$

I mean the statement is true for any points $I_a,I_c$ on $IA,IC$ with condition $\angle I_cBI=\angle IBI_a$, where $I$ is the incenter of $\triangle ABC$. Indeed, $$(C,A,D,AC\cap x)\overset{I}{=}(I_c,I_a,BD\cap I_aI_c,\ell\cap I_aI_c)=-1=(C,A,D,\ell\cap AC),$$where $\ell$ is the external angle bisector of $\angle ABC$ and $x$ is the line connecting $I$ and $\ell\cap I_aI_c$. Thus, $\ell, AC$ and $x$ are concurrent. As $\ell\cap I_aI_c$ lies on $x$, we conclude that $I_aI_c,AC,\ell$ are concurrent. We conclude that $\angle DBQ= 90^\circ$.