In the interior of the square ABCD there is a point P so that <PAD=x, <CDP=45+x. Find ratio DP/AP.
Proof
Clearly <ADP=45-x, so <ADP+<PAD=45, that is, DP passes midpoint E of the outer arc AB of the circle (AB). If F is midpoint of CD, tan(<DEF)=DF/EF=1/3, i.e tan(<DAP)=1/2, that is, A-P-F are collinear. Notice that triangles DAF and PBA are similar, i.e. PB=2AP. Finally notice that triangles DAP and BDP are similar, i.e. AP/PD=PD/PB, hence PD.PD=AP.PB=AP.2AP wherefrom PD=APsqrt(2).
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