Warning: You'll probably need pencil and paper to follow this.
Recently I came across the following problem in a middle/high school geometry textbook:
$\ast$
Suppose $\ QUAD\ $ is a quadrilateral with
$(1) \quad \overline{QU}\cong\overline{AD} \quad ,$
$(2) \quad \overline{UA}\not\cong \overline{DQ} \quad , \quad$ and
$(3) \quad \angle Q\ $ supplementary to $\ \angle A \quad .$
Prove that $\ QUAD\ $ is an isosceles trapezoid.
$\ast$
I first managed to prove the theorem drawing a line through $\ U\ $ parallel to $\ \overline{AD}\ $, then extending $\ \overline{QU}\ $ and $\ \overline{AD}\ $ and working with proportions. Needless to say, the proof got a bit involved.
Later I designed a different proof: Suppose $\ \overline{UA}\ $ and $\ \overline{DQ}\ $ are not parallel; then they meet in a point $\ P\ ,\ $ let's say to the "left" of $\ QUAD\ .\ $ One can now show using $\ (3)\ $ that $\ \triangle PQU \sim \triangle PAD\ ,\ $ which contradicts $\ (1)\ .\ $ So $\ \overline{UA}\ $ is parallel to $\ \overline{DQ}\ ,\ $which basically proves the theorem; property $\ (2)\ $ is there to rule out the case of a rectangle (which I gather is not considered an isosceles trapezoid).
Is there something I'm missing here? Seems a bit involved/indirect for an elementary geometry problem. Thanks!