Given Square ABCD. E,F,G,H are points at AD,AB,BC,and CD respectively. If EG=FH, how to prove that area of EFGH= {(FH)^2}/2 ?

Diagonals are equal, I think, only for a rectangle which includes square or for an isosceles trapezium. With this idea one has to think. I shall submit answer later.

Draw diagram place points as described.

Let the distance of E and G from side DC be e, and g respectively.

Note that EG^2 = a^2 + |e-g|^2 ------------------------------ (1)

Similarly, let the distance of F and H from side AD be f, and respectively.

Note that FH^2 = a^2 + |f-h|^2 ---------------------------------(2)

From(1) and (2) as it is given that EG = FH,

|e-g| = |f-h| --------------------------------- (3)or

(e-g) = -(f-h)-------------------------------- (3a) is also a possibility

Area EFGH, say S =a^2 -{Area(AEF)+Area(BFG)+Area(AEF)+Area(CFH)+Area(DHE)|

=a^2 - (1/2)[{(a-e)*f}+{(a-f)(a-g)}+{(a-h)*g} + (h*e)]

=a^2 - (1/2)[(af-ef)*+(a^2)+ fg-af-ag+ag-hg + (h*e)]

=a^2 - (1/2)[(-ef)+(a^2)+ fg -hg + he]

=a^2 - (1/2)[{(g-e)*f}+(a^2) -{h(g-e)}]

=[(a^2)/2] +(1/2)(e-g)(f-h)

= [(a^2)/2] +(1/2)|(f-h)|^2 [Using (3) and (3A) = [(FH^2)/2]

A●　　　　E●　　　　　　　　D●

　　　　　x　　180-x　　　　90

　　　　　　　　　　　　　　x

　　　　　　　　　　　　　　　　H●

　　　　　　　　I●

F●

B●　　　　　　　G●　　　　　C●

Intersection I

EG = FH ......∠IHD = ∠IEA = x .....EG ⊥ FH

S(EFGH) = FH•EG/2 = (FH)²/2