Team Selection Test

Let $M$ be the intersection of the lines $PE$ and $BC$, $N$ be the intersection of the lines $PF$ and $BC$. We will prove that $MD=ND$.

The power of $B$ with respect to the circumcircle of the triangle $AKD$ gives $$BK\cdot BA=B{D}^2=B{F}^2,i.e.B{K}^2+BK\cdot KF+BK\cdot AF=(BK+KF{)}^2.$$ Therefore, $AF=\frac{KF\cdot BD}{BK}$. On the other hand as $FN\parallel KD$ we have $\frac{ND}{BD}=\frac{KF}{BK}$ and hence $ND=AF$. Similarly we can get $MD=AE$ and then $AE=AF$. Thus, $MD=ND$.

Now note that $\angle DAE=\angle LDC=\angle EMD$ and hence the points $A,E,D,M$ are concyclic. Thus the circumradius of the triangle $EMD$ is $R_2$. Since $PD=2R_4\cdot \sin (\angle PED)$, $MD=2R_2\cdot \sin (\angle MED)$ and $\angle MED=\angle PED$, we get $\frac{PD}{MD}=\frac{R_1}{R_2}$. In a similar way we can get $\frac{R_3}{R_1}=\frac{PD}{ND}$ and the result follows.