Quadrilateral ABCD has points A (0,0) B (a,b) C (a+b,a+b) D (b,a). Prove it is a rhombus for any values of a and b.
I have tried to solve this by saying that since points A and C both have equal x and y coordinates, that this means all 4 sides are equal etc. However, this doesn't sound correct to me. Any ideas?
Define $m(PQ)$ as the slope of PQ.
By inspection, $m(AB)=m(CD)=\dfrac ba$ and $m(BC)=m(AD)=\dfrac ab$, i.e. opposite sides are parallel.
Also, by inspection, $m(AC)\cdot m(BD)=1\cdot (-1)=-1$ i.e. diagonals are perpendicular.
Hence $ABCD$ is a rhombus. $\blacksquare$