ABCDE is a regular pentagon; rays AB and DC intersect at X. Now the area of triangle BCX is 1. What is the area of the pentagon?
I figured out that the area of the pentagon is the square root of 5. (At least that's what I think it is). However, the computation is quite lengthy and involves Angle Bisector Theorem, Ptolemy's Theorem or Similar Triangles.
I'm just wondering if there's a very quick and efficient way to prove this? A bonus would be a cool proof without words, if that exists of course!
Triangle $BXC$ is congruent to triangle $BEC$ and ${A_{BEC}\over A_{ECD}}={BE\over CD}={BE\over BC}$
In triangle $EBC$ we choose a point $Y$ on $BE$ such that $BC=CY$ it is not hard to find that $BC=CY=YE$ and triangles $CBY$ and $EBC$ are similar.
Hence ${BE\over BC}={BC\over BE-BC}\implies BE^2-BE\cdot BC-BC^2=0$
Hence $BE={BC+\sqrt{5}BC\over2}=({1+\sqrt{5}\over2})BC$
So the area of pentagon is $1\cdot(1+2\cdot{2\over 1+\sqrt{5}})=1+({4\over 1+\sqrt{5}})=\sqrt{5}$