kajcha wrote:Since BE = ED, BDE forms a isoc traingle.
extension of CE will cut BD at right angle...
Every line from C will not form a right angle at BD. We have to assume here that CE is the median of the isoc triangle BDE. In that case;
Area BDE = 1/2 x root(2) x (1 + 1/root(2)) = 1/root(2) + 1/2
Arear BEF = 1/2 Area BDE = 1/2 root(2) + 1/4
Area BDC = 1/2 x root(2) x 1/root(2) = 1/2
Now Area BCE = Area BEF - Area BDC
= 1/2 root(2) + 1/4 - 1/2
= 1/2 root(2) - 1/4
= [root(2) - 1]/4
Am I missing anything??
My approach will be right if we consider a 3D figure (which again is an assumption

)
I don't know what to do here.
Also would be great if you can look into
https://www.beatthegmat.com/viewtopic.php?t=5035 and give your inputs.
regards
niks...