Find The Ratio

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Consider the simplified diagram above. Let $H$ be the orthocenter of $\triangle ABC.$ Since $HD \perp BC$ and $HF \perp AB,$ quadrilateral $BFHD$ is cyclic with diameter $BH.$ Note that $\begin{array}{lcl} \angle BFD & = & \angle BHD \\ & = & \angle AHE \\ & = & 90^{\circ} - \angle HAE \\ & = & 90^{\circ} - (90^{\circ} - \angle BCA) \\ & = & \angle BCA. \end{array}$ Similarly, $\angle AFE = \angle BCA.$ Since quadrilateral $ABCP$ is cyclic, $\angle BPA = \angle BCA.$ Let $\angle PBC = \angle PAC= \theta.$ Then, $\angle ABR = \angle ABC - \theta,$ and $\angle BRF = 180^{\circ} - \angle BCA - (\angle ABC - \theta ) = \angle BAC + \theta = \angle FAP.$ Now note that $\angle FRP = 180^{\circ} - \angle BRF = 180^{\circ} - \angle FAP,$ so quadrilateral $AFRP$ is cyclic. We then have $\angle ARP = \angle AFP = \angle BCA = \angle APB,$ so $\triangle ARP$ is isoceles with $AP= AR.$ In a similar manner, we can prove that $AR= AS= AT = AP= AQ= AU.$ Our desired answer is $\dfrac{AR+AS+AT}{AP+AQ+AU} = \boxed{1}.$

This problem is adapted from IMOSL 2010 G1 .