question 4

Question was :-
Let ABCDEF be a convex hexagon in which the diagonals AD, BE, CF are concurrent
at O. Suppose the area of traingle 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.


Solution:
Let OA = a
 OB = b,
 OC = c,
OD = d,
 OE = e,
 OF = f,
 [OAB] =x,
 [OCD] = y,
[OEF] = z,
 [ODE] = u,
[OFA] = v
[OBC]= w.
We are given that v^2 = zx, w^2 = xy and we have to prove that u^2 =yz.
Since∠AOB =∠DOE, we have
u\x =(1/2 de sin∠DOE)/(1/2 ab sin∠AOB)= de/ab

Similarly, we obtain
v/y =fa/cd
w/z =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 (zx)(xy).
This gives u^2=yz, as desired.