Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 3 / 14F / c / Solution

Write $z_j=c+R{e}^{i{\theta }_j}$ with ${\theta }_1<{\theta }_2<{\theta }_3<{\theta }_4<{\theta }_1+2\pi$. In the identitythe phase factors cancel in the cross-ratio. In our convention the two numerator sine factors are negative, while the denominator factors have opposite signs. Thus $r<0$, and the exchanged cross-ratio $1-r$ is greater than one. Consequently $|r|+1=1-r=|1-r|$.

Taking absolute values of the two formulas in part (a) givesMultiplying $|r|+1=|1-r|$ by the common positive denominator proves Ptolemy theorem:The product of the diagonals equals the sum of the products of opposite sides.