Intersecting Triangles

Area ABF = Area ACD + Area FDE - AreaBCE

= (10 * 10)/2 + (10 sin 45 * 10)/2 - ( 20 tan 22.5* 20)/2

= 50 + 35.355339 - 82.842712

= 2.51262

= 2.5126

Visually we see that the areas of the triangles are related as follows $\Delta ABF = \Delta ACD-\Delta EBC+\Delta DFE$

The area of ACD is: $\Delta ACD=\frac{|AC||CD|}{2}=\frac{10^2}{2}=50$

For the area of $DFE$ we need it's height $h_F$ which we can calculate from Pythagoras' theorem:

$h_F^2+h_F^2=|DF|^2 \implies h_F=\sqrt{50}$ Therefore: $\Delta DFE=\frac{h_F |DE|}{2}=\frac{10 h_F}{2}=5\sqrt{50}$

For the area of $EBC$ we need it's height $|BC|$ . If we denote by $F_x$ the point between $CD$ that makes $DFF_x$ a right-angled triangle, then we see that since $EFF_x$ and $CBE$ are congruent it must be the case that

$\frac{|BC|}{|CE|}=\frac{|FF_x|}{|F_xE|}=\frac{h_F}{h_F+|DE|}\implies |BC|=\frac{20h_F}{10+h_F}=\frac{20\sqrt{50}}{10+\sqrt{50}}$ Therefore: $\Delta CBE=\frac{|BC||CE|}{2}=10|BC|=\frac{200\sqrt{50}}{10+\sqrt{50}}$

Substituting the initial equation, we get:

$\Delta ABF = 50- \frac{200\sqrt{50}}{10+\sqrt{50}} +5\sqrt{50} \approx 2.513$