[asy][asy] size(7cm); pointpen=black; pathpen=black; pair A=dir(120), B=dir(-160), C=dir(-20), D=foot(A, B, C), F=incenter(A, D, C), E=extension(A, F, D, C); D(A--B--C--cycle, red); DPA(D--A--E^^B--F, orange); markscalefactor=0.01; DPA(rightanglemark(A, D, B)^^rightanglemark(A, F, B), blue); DPA(MA(D, A, E, 0.28, darkgreen)^^MA(E, A, C, 0.22, darkgreen)); D("A", A, A); D("B", B, B); D("C", C, C); D("D", D, S); D("E", E, S); D("F", F, NE); [/asy][/asy] Since $\angle BDA=\angle BFA=90^\circ\implies ABDF$ is cyclic, so $$\angle CDF=\angle BAE=45^\circ=\frac{\angle CDA}{2}$$thus $F$ is the incenter of $\triangle CDA$. Therefore, $$\angle BFC=\angle BFE+\angle EFC=90^\circ+\left(90^\circ-\frac{\angle CDA}{2}\right)=\boxed{135^\circ}$$

Simple but nice Let $H = AD \cap BF$ and let $\angle CAE = \angle EAD = x$. $\angle FBD = \angle FAD = x$ since $AFDB$ is cyclic which means $\angle ADF = \angle ABF = \angle ABD - \angle FBD = 90 - \angle BAD - \angle FAD = 90 - (45-x) - (x) = 45^\circ$ which also means $\angle FDC = 45^\circ$ which means that $F$ must be the incenter of $\triangle ADC$. Since $\angle BCA = 180 - (\angle BAC + \angle ABC) = 180 - (45+x + 45+x) = 90 - 2x$, we know that $\angle FCB = 45-x$. So, $\angle BFC = 180 - (\angle FBC + \angle FCB) = 180 - (x + 45 - x) = 135^\circ$

angle BAE=45° angle ABD=45° let angle DAE= angle EAC=x angle BAD=45°-x angle ABD=45°+x= angle BAC So, AC=BC let angle CFE=y In ∆BFC, sin(90°+y)/sin x=BC/ FC cos y/ sin x=BC/ FC In ∆ AFC, sin y/ sin x=AC/ FC so, cos y=sin y y =45° angle BFC=90°+45°=135° @ KRISHIJIVI

Let $K=BF \cap AC$. Now notice that $\triangle ABF$ is isosceles so $AF=BF$ and $\angle FAB=\angle FBA=45^\circ$ and $\angle DAE=\angle FBE=\angle CAE$ therefore $\angle CAE+\angle BAE=\angle ABK+ \angle CBK$ $\implies$ $\angle CAB=\angle CBA$ which means $AB=AC$. Notice that $\triangle AFK \cong \triangle BFE (A.S.A)$ $\implies$ $AK=BE$, $KC=EC$ and $FK=FE$ so $FKEC$ is a kite therefore $CF$ is the angle bissector of $\angle KCE$ so $\angle CFE=\angle CFK=45^\circ$. Finally $\angle BFC=90^\circ+45^\circ=135^\circ$

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