Isosceles trapezoid \(ABCD\) has parallel sides \(AD\) and \(BC\), with \(BC<AD\) and \(AB=CD\). There is a point \(P\) in the plane such that \(PA=1\), \(PB=2\), \(PC=3\), and \(PD=4\). What is \(\frac{BC}{AD}\)?
Solution
Answer: \(\frac13\)
Reflect \(P\) over the perpendicular bisector of \(BC\) to obtain \(P’\). Then \(P’A=PD=4\), \(P’D=PA=1\), \(P’C=PB=2\), and \(P’B=PC=3\). The isosceles trapezoids \(DAPP’\) and \(CBPP’\) are cyclic.
Apply Ptolemy’s theorem to each:
\[PP’\cdot AD+1\cdot1=4\cdot4,\qquad PP’\cdot BC+2\cdot2=3\cdot3.\]
Thus \(PP’\cdot AD=15\) and \(PP’\cdot BC=5\). Dividing gives \(\frac{BC}{AD}=\frac5{15}=\frac13\).