Let $F'$ be the reflection of $D$ in $A$ and $B'$ be the point on $BC$ such that $B'F'$ is tangent to the circle with center $A$ and radius $AD$. [asy][asy]import graph; size(9.21cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -3.27, xmax = 5.94, ymin = -6.24, ymax = 0.45; /* image dimensions */ /* draw figures */ draw(circle((-0.06,-2.54), 2.63)); draw((-0.06,-2.54)--(2.64,-2.33)); draw(circle((2.64,-2.33), 2.62)); draw((1.82,0.16)--(0.4,0.05)); draw((0.4,0.05)--(2.44,-1.72)); draw((1.82,0.16)--(0.06,-1.87)); draw((0.06,-1.87)--(2.64,-2.33)); draw((3.82,-5.9)--(-0.52,-5.12)); draw((2.44,-1.72)--(-0.52,-5.12)); draw((0.4,0.05)--(-0.52,-5.12)); draw((1.82,0.16)--(3.82,-5.9)); /* dots and labels */ dot((-0.06,-2.54),dotstyle); label("$A$", (-0.48,-2.55), NE * labelscalefactor); dot((0.4,0.05),dotstyle); label("$D$", (0.02,-0.35), NE * labelscalefactor); dot((2.44,-1.72),dotstyle); label("$E$", (2.68,-1.59), NE * labelscalefactor); dot((2.64,-2.33),dotstyle); label("$B$", (2.9,-2.26), NE * labelscalefactor); dot((0.06,-1.87),dotstyle); label("$F$", (-0.36,-1.88), NE * labelscalefactor); dot((1.82,0.16),dotstyle); label("$C$", (2.21,-0.11), NE * labelscalefactor); dot((-0.52,-5.12),dotstyle); label("$F'$", (-0.95,-4.76), NE * labelscalefactor); dot((3.82,-5.9),dotstyle); label("$B'$", (3.99,-5.67), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Now $BF\parallel B'F'$ implies $\angle CBF=\angle EB'F$. Since $\triangle BCF$ and $\triangle B'EF'$ are isosceles, we get $\angle FCB=\angle F'EB'$, so $CF\parallel EF'$. Since $DF'$ is a diameter, $DE\perp EF'$, so $CF\parallel EF'\perp DE\implies CF\perp DE$.

Let $G = CF \cap DE$. Since $ BF \perp AD$ and $AE \perp BC$, then $AFBE$ is cyclic. Notice that \[ 2 \angle DFC = \angle FBC = \angle FBE = \angle DAE = 2 \angle DEC\]so $\angle DFC = \angle DEC$. Finally \[ \angle ECG = \angle BCF = 90^{\circ} - \angle DFC = 90^{\circ} - \angle DEC = 90^{\circ} - \angle GEC\]then $\angle EGC = 90^{\circ}$ and thus $DE \perp CF$ $\blacksquare$

oo no new points introduced here.. Note that $BF\perp AF, AE\perp BE\implies AFEB \text{ is cyclic}$. Note that $$\angle FCE=\angle CFB=90-\frac{\angle FBE}{2}=90-\frac{\angle DAE}{2}=\angle FDE\implies DCEF\text{ is cyclic}.$$To finish, $$\angle DEC=180-(\angle DEA+\angle AFD)=180-(\angle ADE+90)=90-\angle ADE=90-\angle FCE\implies DE\perp FC.$$