Problem 2

Let $F$ be the foot of the altitude from $C$ to $AB$, and let $C'$ be the reflection of $C$ through the point $F$; let $L$ be the line through $C'$ parallel to $AB$, so $L$ lies at distance $CF$ from $AB$, on the opposite side from $C$. For any point $D$ on $AB$, if line $CD$ meets $L$ at a point $E'$, then $D$ lies exactly midway (in the direction perpendicular to $AB$) between $C$ and $E'$, which gives $DE'=DC$. So the condition $DC=DE$ from Step 1 says exactly that the second intersection $E$ of line $CD$ with the circumcircle coincides with this point $E'$ — in other words, that $E$ lies on the line $L$.