Question

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

• A. AP
• B. HP
• C. GP
• D. None of these

Answer 1

Answer: (C)

Let $a,b,c,d,e,f$ be the length of the sides $OA,OB,\dots$
respectively and $x,w,y,u,z,v$ be this area respectively.

So, $\frac{u}{x}=\frac{\frac{1}{2}\text{ de }\sin \angle DOE}{\frac{1}{2}ab\sin \angle AOB}=\frac{de}{ab}$
Similarly; $\frac{v}{y}=\frac{fa}{cd},\frac{w}{z}=\frac{bc}{ef}$
On multiplication $x^2{y}^2{z}^2=u^2{v}^2{w}^2=u^2(zx)(xy)$
So, $u^2=yz$