A, B, C and D are non-coplanar points. ∠ABC = ∠ADC and ∠BAD = ∠BCD. Show that AB = CD and BC = AD.
Solution
It is worth looking first at the coplanar case. If ABCD is convex, then opposite angles are equal, so it is a parallelogram and opposite sides are equal. In this case the result is true. But if ABDC is convex, then ∠ABC = ∠ADC implies that A, B, D, C lie on a circle and hence that ∠BAD = ∠BCD. In this case it is certainly not true that AB = CD or that BC = AD.
Applying the cosine formula to angles ABC and ADC gives: (AB2 + BC2 - AC2)/(2 AB BC) = (AD2 + CD2 - AC2)/(2 AD CD). Hence (AB·CD - AD·BC)(AB·AD - BC·CD) + AC2(AB.BC - AD.CD) = 0 (*). Similarly, applying it to the other two angles gives: (BC2 + CD2 - BD2)/(2 BC CD) = (AB2 + AD2 - BD2)/(2 AB·AD) and hence (AD·BC - AB·CD)(AB·BC - CD·AD) + BD2(BC·CD - AB·AD) = 0 (**).
Put x = (AB·CD - AD·BC), y = (AB·AD - BC·CD), z = (AB·BC - AD·CD). Then (*) and (**) become xy + AC2z = 0, xz + BD2y = 0. So if either of y or z is 0, then the other is also. But if y and z are 0, then AB/CD = BC/AD = AD/BC, so BC = AD and hence also AB = CD. On the other hand, if neither y nor z is 0, then we can deduce that AC2BD2 = x2, and hence AC·BD = ±x = ±(AB·CD - AD·BC). Thus either AC·BD + AD·BC = AB·CD or AC·BD + AB·CD = AD·BC.
In either case we can use Ptolemy's theorem which tells us that A, B, C, D must be (1) coplanar, and (2) concyclic (or collinear). But we are told that they are not coplanar. Hence BC = AD and AB = CD as required.
© John Scholes
jscholes@kalva.demon.co.uk
27 Jan 2001