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 $OA , OB , \ldots$
respectively and $x , w , y , u , z , v$ be this area respectively.

So, $\frac{u}{x}=\frac{\frac{1}{2} \text { de } \sin \angle DOE }{\frac{1}{2} ab \sin \angle AOB }=\frac{d e}{a b}$
Similarly; $\frac{v}{y}=\frac{f a}{c d}, \frac{w}{z}=\frac{b c}{e f}$
On multiplication $x^{2} y^{2} z^{2}=u^{2} v^{2} w^{2}=u^{2}(z x)(x y)$
So, $u^{2}=y z$