Vectors in trapezoid

Question

Let $ABCD$ be a trapezoid having the bases $AB$ and $CD$ with $AB>CD$. We know that $AD=DC=CB$. Express the vector $\vec{CD}$ in terms of $\vec{AD}$ and $\vec{BD}$. It is clear that $\vec{CD}=k\vec{AB}$, where $k\in\mathbb{R}\setminus\lbrace 0\rbrace$. Also, it is easy to express $\vec{AB}$ in terms of $\vec{AD}$ and $\vec{BD}$. Is there a way to find $k$? For sure the condition $AD=DC=CB$ has to be used in some way.

Answer 1

Observe that your condition $AD=DC=CB$ defines the shape of your trapezoid (I mean, it is determined up to similarity). Consider now a regular hexagon and cut it in half by a diagonal joining opposite vertices. Each of its halves is a trapezoid with your property, so the proportions of the sides in your trapezoid are exactly the proportions in that half-hexagon. The side of a regular hexagon is equal to the radius of the circumscribed circle, that is $CD=AB/2$. So your $k=1/2$.