Hello,
For that, I will use a rough method, using more algebra than geometry.
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Because I don't remember the formula for $\displaystyle \arctan a+\arctan b$, I'll show it.
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For that I will need to find the formula (which I don't remember either) of tan(c+d) :
$\displaystyle \tan(c+d)=\frac{\sin(c+d)}{\cos(c+d)}$
$\displaystyle \tan(c+d)=\frac{\sin(c)\cos(d)+\cos(c)\sin(d)}{\co s(c)\cos(d)-\sin(c)\sin(d)}$
Let's multiply by $\displaystyle \frac{~~~~\frac{1}{cos(c)\cos(d)}~~~~}{~~~~\frac{1 }{cos(c)\cos(d)}~~~~}$
This gives :
$\displaystyle \tan(c+d)=\frac{\tfrac{\sin(c)}{\cos(c)}+\tfrac{\s in(d)}{\cos(d)}}{1-\tfrac{\sin(c)}{\cos(c)} \cdot \tfrac{\sin(d)}{\cos(d)}}$
$\displaystyle \boxed{\tan(c+d)=\frac{\tan(c)+\tan(d)}{1-\tan(c)\tan(d)}}$
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Therefore, we have :
$\displaystyle \tan(\arctan(a)+\arctan(b))=\frac{\tan(\arctan(a)) +\tan(\arctan(b))}{1-\tan(\arctan(a))\tan(\arctan(b))}$
$\displaystyle \tan(\arctan(a)+\arctan(b))=\frac{a+b}{1-ab}$
$\displaystyle \implies \boxed{\arctan(a)+\arctan(b)=\arctan\left(\frac{a+ b}{1-ab}\right)}$
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Now let's go back to the problem.
So we have $\displaystyle \angle DPC+\angle DBC=\arctan(1/3)+\arctan(1/2)$
$\displaystyle \arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{1/3+1/2}{1-(1/2)*(1/3)}\right)$ (formula above).
$\displaystyle \arctan(1/3)+\arctan(1/2)=\arctan\left(\frac{5/6}{1-1/6}\right)=\boxed{\arctan(1)}$
Therefore, $\displaystyle \boxed{\angle DPC+\angle DBC=\angle DQC}$