48th International Mathematical Olympiad Vietnam 2007 Shortlisted Problems with Solutions

Let $t=\frac{AD}{BC}$. Consider the homothety $h$ with center $P$ and scale $-t$. Triangles $PDA$ and $PBC$ are similar with ratio $t$, hence $h(B)=D$ and $h(C)=A$.



Let $Q^{\prime }=h(Q)$ (see Figure 1). Then points $Q$, $P$ and $Q^{\prime }$ are obviously collinear. Points $Q$ and $P$ lie on the same side of $AD$, as well as on the same side of $BC$; hence $Q^{\prime }$ and $P$ are also on the same side of $h(BC)=AD$, and therefore $Q$ and $Q^{\prime }$ are on the same side of $AD$. Moreover, points $Q$ and $C$ are on the same side of $BD$, while $Q^{\prime }$ and $A$ are on the opposite side (see Figure above).

By the homothety, $\angle A{Q}^{\prime }D=\angle CQB=\angle AQD$, hence quadrilateral $A{Q}^{\prime }QD$ is cyclic. Then $$\angle DAQ=\angle D{Q}^{\prime }Q=\angle D{Q}^{\prime }P=\angle BQP$$ (the latter equality is valid by the homothety again).