Question

let ABCD BE A SQUARE OF SIDE LENGTHS 2, and let E and F be MIDPOINTS OF AD and CD respectively. let G and H
be a points where AF intersects EB and BD respectively what is the area of i) triangle AGE,and quadrilateral DEGH

Answer 1

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