# Area of Pentagon using geometry

A given convex pentagon $ABCDE$ has the property that the area of each of the five triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$ is unity. calculate the area of the pentagon $($which need not be regular$)$

Found the solution at this link http://artofproblemsolving.com/wiki/index.php/1972_USAMO_Problems/Problem_5 but dint understand any part of it. Help me !

I'll approach the question in a constructive way. If we start with an arbitrary side $AB$, the altitude of both triangles $ABC$ and $ABE$ with respect to base $AB$ must be $h=2/AB$: thus $C$ and $E$ must belong to a line parallel to $AB$, at a distance $h$ from $AB$.
Let then $D$ be the intersection of lines $AD$ and $BD$, parallel respectively to $BC$ and $AE$. By construction, triangles $BCD$ and $AED$ also have unit area. But triangle $CDE$ has not, in general, unit area: that happens only for some particular positions of points $C$ and $E$.
Let $K$, $G$ and $H$ the projections of $C$, $D$ and $E$ on line $AB$, and suppose triangle $CDE$ has unit area, so that its altitude with respect to base $EC$ is $h'=2/EC=2/HK$. By similar triangles we have $CK:DG=BK:AG=AH:BG$, that is: $$CK:DG=(BK-AH):(AG-BG), \quad\hbox{or:}\quad h:(h+h')=(HK-AB):AB.$$ Plugging there $h=2/AB$ and $h'=2/HK$ we can solve for $HK/AB$, which is the same as $EC/AB$, and find: $${EC\over AB}={1+\sqrt5\over2}.$$ This is then the condition to be respected for the pentagon to have the desired properties. We are free to set $A$, $B$ and $C$, but $E$ is then fixed by this condition and $D$ is fixed by the above construction.
Let's now compute the area of pentagon $ABCDE$ thus constructed. It can be decomposed as the sum of trapezoid $ABCE$ and triangle $CDE$, thus: $$Area_{ABCDE}=Area_{CDE}+Area_{ABCE}=1+{1\over2}(AB+EC)h= 1+{1\over2}(AB+EC){2\over AB}=2+{EC\over AB}.$$ That shows that all pentagons have indeed the same area and that $$Area_{ABCDE}=2+{1+\sqrt5\over2}.$$