Question

Let $ABCDEF$ be a convex hexagon in which the diagonals $AD,BE,CF$ are concurrent at $O.$ Suppose the area of triangle $OAF$ is the geometric mean of those of $OAB$ and $OEF$; and the area of triangle $OBC$ is the geometric mean of those of $OAB$ and $OCD.$ Prove that the area of triangle $OED$ is the geometric mean of those of $OCD$ and $OEF.$

Answer 1

Let $OA=a,OB=b,OC=c,OD=d,OE=e,OF=f,\left[OAB\right]=x,\left[OCD\right]=y,\left[OEF\right]=z,\left[ODE\right]=u,\left[OFA\right]=v$

and $\left[OBC\right]=w$. We are given that $v^2=zx,w^2=xy$ and we have to prove that $u^2=yz$.
Since $\angle AOB=\angle DOE$, we have
$\frac{v}{y}=\frac{\frac{1}{2}desin\angle DOE}{\frac{1}{2}absin\angle AOB}=\frac{de}{ab}$
Similarly, we obtain $\frac{v}{y}=\frac{fa}{cd},\frac{w}{z}=\frac{bc}{ef}$
Multiplying, these three equalities, we get $uvw=xyz$. Hence
$x^2{y}^2{z}^2=u^2{v}^2{w}^2=u^2\left(zx\right)\left(xy\right)$.
This gives $u^2=yz$, as desired.