See the diagram.

We prove this by using the similar triangles principles.

Let angle ABE = x. Also, angle DAF = x by similarity.

=> angle GAB = 90° - x

=> In triangle AGB, angle AGB = 90°.

=> angle AGE = 90°.

AF = √(AD²+DF²) = √(2²+1²) = √5.

Right angle triangles ADF and AGE are similar, as two angles are equal.

=> AG / AD = EG / DF = AE / AF

=> AG / 2 = EG / 1 = 1 / √5

=> AG = 2/√5 and EG = 1/√5

Area of triangle AGE = 1/2 * AG * EG = 1/2 * 2/√5 * 1/√5* = 1/5*

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Draw a perpendicular from H onto DF. As HI || AD, angles FAD = angle FHI = x. Hence the triangles HIF and FAD are similar.

=> HI / AD = IF / DF = HF / AF

=> HI / 2 = IF / 1 = HF / √5

=> HI = 2 IF

In the triangle DHI, angle HDI = 45°, as the diagonal bisects angle D. Since angle DIH = 90°, then angle DHI = 45°. Hence, it is isosceles and DI = HI.

DF = 1

= DI + IF = 2 IF + IF = 3 IF

=> IF = 1/3 => HI = 2/3

Area of triangle ADF = 1/2 * AD * DF = 1

Area of triangle DHF = 1/2 * HI * DF = 1/2 * 2/3 * 1 = 1/3

Area of triangle AEG = 1/2 * AG * EG = 1/2 * 2/sqroot(5) * 1/sqroot(5) = 1/5

Finally, *the area of quadrilateral DEGH*

= area triangle ADF - area triangle AEG - area trianlge DHF

= 1 - 1/5 - 1/3 = *7/15*