The question shows three triangles (all right angled triangles).
Two of them are inscribed inside a bigger one.
Triangle DEB and triangle BEF are both inscribed inside the larger triangle DEF.
We shall use the ratios of these triangles to calculate the unknown sides;
[tex]\begin{gathered} \frac{DB}{ED}=\frac{DE}{DF} \\ \frac{DB}{4}=\frac{4}{(DB+4+DB)} \\ \frac{DB}{4}=\frac{4}{4+2DB} \\ \text{Cross multiply and you'll have,} \\ DB(4+2DB)=16 \\ 4DB+2DB^2=16 \\ Factorize\text{ the left hand side} \\ 2(2DB+DB^2)=16 \\ \text{Divide both sides by 2} \\ 2DB+DB^2=8 \\ \text{Subtract 8 from both sides} \\ DB^2+2DB-8=0 \\ We\text{ now have a quadratic equation, and the factors are,} \\ (DB+4)(DB-2)=0 \\ \text{Therefore, } \\ DB+4=0 \\ DB=-4 \\ OR,DB-2=0 \\ DB=2 \end{gathered}[/tex]The length of DB is 2
