Question

Two Chords AB and CD of a cirle intersect at E such that AE = 2•4cm, BE = 3•2cm And CE = 1•6cm. The Length of DE is:

Answer 1

if these chord are interesting then
AE/BE = CE/DE
2.4/3.2 = 1.6/DE
DE = 1.2 cm

hope it help...TY

Answer 2

$\frac{ae}{be} = \frac{ce}{de} \\ \frac{2.4}{3.2} = \frac{1.6}{x} \\ x = 1.2 \: cm$
hope this helps