The existence of $E$ is equivalent to $\angle DAB=\angle DAC$, so $AO_2$ bisects $\angle BAC$. Thus $AC=AB$, since this bisector passes through the circumcenter of $\triangle ABC$. If we invert about $A$, (marking points in the newly inverted diagram with $'$s) we get that $D'A'$ is the perpendicular bisector of $B'C'$, and $C'A'\perp B'D'$, so $A'$ is the orthocenter of isosceles $B'D'C'$. So additionally $D'$ is the orthocenter of $A'B'C'$, which means if we invert about $D'$, in the newest diagram $D"$ (we mark points in the twice inverted diagram with $"$s) becomes the excenter of $B"A"C"$. So since $B"A"C"$ is isosceles, the midpoint $X$ of minor arc $B"C"$ is the center of $(D"B"C"")$, but then $XB"\perp B"A"$ and $XC"\perp C"A"$. Thus $A"B"$ is tangent to $(D"B"C")$, and $A"$ is be the intersection of the tangents from $B",C"$ to $(D"B"C")$. Thus inverting back through $D"$, $A'$ is the intersection of the circles through $D'$ tangent to $B'C'$ through $B',C'$, and then inverting back into the first diagram through $A'$,$DC$ is tangent to $(ABC)$.

Note that $\angle AO_2O_1=\angle BO_2O_1=90^{\circ}-\angle BED$. Again, $\angle ABC=\dfrac{1}{2}\angle AO_2C=90^{\circ}-\angle O_2AC=90^{\circ}-\angle O_1AD=90^{\circ}-\angle BED$ since $AO_1\parallel DE$ and $AD\parallel BE$. So $\angle ABC=\angle ACB\Rightarrow AC=AB$. Thus $\angle AO_2C=\angle AO_2B$. This means $\angle BAO_2=\angle CAO_2=\angle O_1AD=\angle O_1DA=\angle O_1DO_2\Rightarrow DO_1\parallel AB$. But $AB\perp O_1O_2$ so $DO_1\perp O_1O_2$. Now $\angle O_2CA=\angle O_2AC=\angle O_1AD=\angle O_1DA$ i.e $\angle O_1CO_2=\angle O_1DO_2$ which means $O_1DCO_2$ is cyclic. So, $\angle O_2CD=180^{\circ}-\angle O_2O_1D=180^{\circ}-90^{\circ}=90^{\circ}$ since $DO_1\perp O_1O_2$. Thus $DC\perp CO_2$ as desired.

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My solution for this nice problem: We must to prove that $\angle DAC=2\angle AO_2C.$We know $ABED$ and $EDCA$ are isoceles trapezoid and parallelogram, respectively.Then $AB=DE=AC.$ Then we can find easily $\angle O_2AB=\angle O_2CA=\angle O_2AC=\angle DAO_1=\angle ADO_1,$ then $O_1DCO_2$ is cyclic.Also we know $\angle DCA=\angle DEA=2\angle DO_1C=2\angle DO_2A=2\angle AO_2C.$ As desired.

First Note that ABED is isosceles trapezoid. Let DO1 meet $\odot O_1$ at S. ∠O2CA = ∠O2AC = ∠O1AD so we need to prove ∠O1AS = ∠DCA. ∠O1AS = ∠ASO1 = ∠AED so we need to prove DCAE is parallelogram. DE || AC and ∠CAO2 = ∠O1AD = ∠ABE = ∠BAO2 so CA = BA = DE so DCAE is parallelogram. we're Done.

As shown in the figure, $\odot O_1$ and $\odot O_2$ intersect at two points $A$, $B$. The extension of $O_1A$ meets $\odot O_2$ at $C$, and the extension of $O_2A$ meets $\odot O_1$ at $D$, and through $B$ draw $BE \parallel O_2A$ intersecting $\odot O_1$ again at $E$. If $DE \parallel O_1A$, prove that $DC \perp CO_2$.