Points D, E and F are the midpoints of sides AB, BC and CA of \(\Delta ABC\). Show that \(\Delta DEF \sim \Delta CAB\) and find \(A(\Delta DEF) : A(\Delta ABC)\).

Answer: \(\Delta DEF \sim \Delta CAB\) by the SSS test, and \(A(\Delta DEF) : A(\Delta ABC) = 1 : 4\)

[Figure: \(\Delta ABC\) with midpoints D on AB, E on BC, F on CA joined to form \(\Delta DEF\).]

By the midpoint theorem, a segment joining the midpoints of two sides of a triangle is half the third side:

D, E midpoints of AB, BC: \(DE = \dfrac{1}{2} AC\). E, F midpoints of BC, CA: \(EF = \dfrac{1}{2} AB\). F, D midpoints of CA, AB: \(FD = \dfrac{1}{2} BC\).

\(\therefore \dfrac{DE}{CA} = \dfrac{EF}{AB} = \dfrac{FD}{BC} = \dfrac{1}{2}\).

So, in the correspondence \(DEF \leftrightarrow CAB\) (D with C, E with A, F with B), all three pairs of sides are in the same ratio. \(\therefore \Delta DEF \sim \Delta CAB\) (SSS test of similarity).

By the theorem of areas of similar triangles, \(\dfrac{A(\Delta DEF)}{A(\Delta CAB)} = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}\). So \(A(\Delta DEF) : A(\Delta ABC) = 1 : 4\).