Find the length of perpendicular DF. The question is from the topic Similarity of Triangles. It goes like this:
$ABCD$ is a quadrilateral with $\angle A= \angle C = 90°$. AE = 5, BE = 12 and AC = 21. Find the length of DF.
The figure:-

My attempt:- 
AB = 13 (Pythagoras Theorem)
EC = AC - AE = 16
BC = 20 (Pythagoras Theorem)
Let EF = a, so FC = 16 - a 
Let AD = z and CD = y 
Let DF = x 
(5 + a)² + x² = z² 
25 + a² + 10a + x² = z² ----(1)
x² + (16 - a)² = y²
256 + a² - 32a + x² = y² ----(2) 
z² + 169 = BD² = y² + 400
z² - y² = 231
Substituting (1) and (2):- 
25 + a² + 10a + x² - x² -256 - a² + 32a = 231 
42a = 462
$\therefore a = 11$
EF = a = 11 and FC = 16 - a = 5 
I don't know what to do now.
 A: Extend $BE$ to intersect $AD$ at point $G$. $\triangle{ABE} \sim \triangle{AGE}$, $\frac{AE}{EG}=\frac{BE}{AE}, EG=\frac{25}{12}$. 
$\triangle{AEG} \sim \triangle{AFD}$, $\frac{AE}{EG}=\frac{AF}{FD}, FD=\frac{16 \cdot 25}{12 \cdot 5}=\frac{20}{3}$.
A: I must applaud your attempt as pretty ingenious, but you mention that you have found this problem under the chapter called “Similarity”, so we must finish the problem by establishing a similarity of two triangles.
You should be able to see, by a simple angle-chasing argument, and the utilisation of the fact that $\angle C=90^{\circ}$, that $\angle BCE=\angle FDC$ and thus $\triangle BEC\sim\triangle CFD$. This gives, $$\frac{BE}{CF}=\frac{EC}{FD}$$
so $FD=\dfrac{EC\cdot CF}{BE}=\dfrac{20}{3}.$
A: Notice that the angles $\angle{BCE}$ and $\angle{CDF}$ are equal as:-
$$\angle{BCE}+\angle{FCD} = 90°      \text{[given]}$$
$$\angle{CDF}+\angle{FCD} = 90°  \text{[ this follows from the angle sum property of  } \bigtriangleup{CFD}\text{ ]}$$
Subtracting them we get $\angle{BCE} = \angle{CDF}$ 
Now that $ \bigtriangleup{CFD} \sim \bigtriangleup{BEC}$ by $AA$ similarity. This is because they have 2 right angles and 2 angles ($\angle{BCE}$ and $\angle{CDF}$) equal.Thus:-
$$\frac{CF}{BE} = \frac{FD}{EC}$$
$$\frac{5}{12} = \frac{FD}{16}$$
$$DF = \frac{20}{3} = 6.67$$
