AD, EF and BD are perpendicular to AB.
AD = 3 and BC = 5
What is the length h?
As triangles ABD and DBF are similar:Originally Posted by Natasha1
$\displaystyle
\frac{h}{3}=\frac{DB}{AB}\ \ \ \dots (1)
$
Also trianglea ABC and ADF are similar:
$\displaystyle
\frac{h}{5}=\frac{AB-DB}{AB}
$
So dividing these:
$\displaystyle
\frac{5}{3}=\frac{DB}{AB}\ \frac{AB}{AB-DB}=\frac{DB}{AB-DB}
$
Hence:
$\displaystyle
DB=\frac{5}{8}AB
$
Substituting this into $\displaystyle (1)$ then gives:
$\displaystyle
\frac{h}{3}=\frac{5}{8}
$,
or:
$\displaystyle
h=\frac{3\times 5}{8}=\frac{15}{8}
$
RonL