The purple point

As $BD=DC=7$ , by pythagoras theorem $AC^2=AD^2+DC^2$ so $AD=\sqrt{AC^2-DC^2}=\sqrt{25^2-7^2}=24 \therefore AE=ED=12$ .

Then, using angles, we get the next and conclude $\angle BDF=\angle AEF$

Now, we have to prove that $\triangle BDF \sim \triangle AEF$ , that means $\frac{BD}{AE}=\frac{DF}{EF}$

$\triangle EDB \sim \triangle EFD \Rightarrow \frac{EF}{ED}=\frac{FD}{DB}$ as $ED=AE \therefore \frac{FD}{DB}=\frac{EF}{AF} \Rightarrow \frac{BD}{AE}=\frac{FD}{EF}\blacksquare$

Effectively $\triangle BDF \sim \triangle AEF \Rightarrow \boxed{\frac{AF}{BF}=\frac{AE}{BD}=\frac{12}{7} \therefore a+b= 19}$

We can solve this problem by using cos rule to find AF and BF in ΔBCF, and ΔAEF respectively.Before that, we need to find EF and FC. Apply Pythagoras in ΔEDC: EC = √193. To find FC we use similarity between ΔEDC and ΔDCF, FC/DC = DC/EC, so FC=49/√193, We can find EF=EC – FC = 144/√193. We can use cos rule to find BF, with cos(BCF) =DC/EC = 7/√193, BF^2 = BD^2 + FC^2 – 2 BD. FC.cos(BCF) = (49)(625)/193. BF=175/√193. By the same way, with cos (AEF) = - cos(DEC) = -12/√193, AF^2 = AE^2 + EF^2 + 2 AE.EF.cos(DEC) = (90 000)/193. AF=300/√193. So, AF/BF = 300/175 = 12/7, or a+b=19