Denote the angles $\angle{BAC}=\alpha$, $\angle{ABC}=\beta$, $\angle{BCA}=\gamma$. Note that the condition on the angles is equivalent to $2\angle{ADB} + \angle{DAB} = 180$. Let's find angle $\angle DAB$: \begin{align*} 2\angle{ADB} + \angle{DAB} = 180\\ \Rightarrow 2\angle{ADB} + \angle{CAD} - \angle{CAO} = 180\\ \Rightarrow 2(180-\beta -\angle{DAB}) + \angle{CAD} - (90-\beta) = 180\\ \Rightarrow \angle{CAD} - 2\angle{DAB} = 180 - \beta + 90 + 2\beta - 360\\ \Rightarrow \angle{CAD} - 2\angle{DAB} = \beta - 90\\ \Rightarrow \alpha - \angle{DAB} - 2\angle{DAB} = \beta - 90\\ \Rightarrow \angle{DAB} = 30 + \frac{\alpha-\beta}{3}. \end{align*} Similarly, $\angle{DAC} = 30 + \frac{\alpha-\gamma}{3}$. Now let's find $\angle{ADB}$: \begin{align*} \angle{DAB} + \beta + \angle{BDA}=180\\ \Rightarrow 30 + \frac{\alpha + 2\beta}{3} + \angle{BDA} = 180\\ \Rightarrow\angle{BDA} = 150 - \frac{\alpha + 2\beta}{3} = 90 + \frac{\gamma-\beta}{3}\\ \end{align*} Now let's find $MBA$ and $MCA$. Remember $P$, $Q$ are the circumcenters of $\triangle{ADB}$ y $\triangle{ADC}$ respectively: \begin{align*} \angle{MBA}=90-\angle{ADB}=\frac{\beta - \gamma}{3}\\ \angle{MCA} = \angle{ADC} - 90 = (180 - \angle{ADB}) - 90 = \frac{\beta - \gamma}{3}.\\ \end{align*} Note that $\angle{MBA}=\angle{MCA} \Rightarrow ABCM$ is cyclical. Let $X=AM \cap BC$. Let's find angle $\angle{XAD}$: \begin{align*} \angle{XAD} = 180-\angle{MAD} = 180 -\angle{MAB} + \angle{DAB}\\ = \angle{MCD} + \angle{DAB} = \angle{QCA} + \gamma + \angle{DAB}\\ = 30 + \frac{2\gamma + \alpha}{3} = 90 + \frac{\gamma - \beta}{3}. \end{align*} Note that $\angle{XAD} = \angle{BDA}$, the triangle $\triangle{AXD}$ is isosceles and its prependicular bisector $PQ$ is the bisector of angle $\angle{AXD}$. $\blacksquare$

An alternate solution. Firstly as above ,we angle chase like above to get $ABCM$ as cyclic . Then using the given condition , we get quadrilaterals $AMPQ$ and $BCPQ$ as cyclic . Then radical axis on $(AMPQ),(BCPQ),(ABC)$ just finishes it.

Nice Claim1 : $AMCB$ is cyclic. Proof : we have $\angle PBD = \angle 90 - \angle DAB$ and $\angle QCD = \angle 90 - \angle DAC$ so $\angle BMC = \angle DAB + DAC = \angle BAC$. Claim2 : $BPQC$ is cyclic. Proof : $\angle BPQ = \angle BPD + \angle DPQ = 2\angle DAB + \angle ABD$ and $QCD = \angle 90 - \angle DAC$ so we need to prove $\angle 2DAB + \angle ABD = \angle 90 + \angle DAC$. we have $\angle ADC = \angle ADB + \angle DAO$ so $\angle DAB + \angle ABC = \angle 180 - \angle DAB - \angle ABC + \angle DAC - \angle 90 + \angle ABC = \angle 90 - \angle DAB + \angle DAC \implies\angle 2DAB + \angle ABD = \angle 90 + \angle DAC$. Claim3 : $AMQP$ is cyclic. By simple angle chasing we have $APB$ and $AQC$ are similar so $\angle DAP = \angle CAQ \implies \angle PAQ = \angle BAC = \angle PMQ$ Now By Radical Axis Theorem on $AMCB$, $AMQP$ and $BPQC$ we have $AM$, $PQ$ and $BC$ are concurrent.