Divide the Land

Problem

Idea : Given a trapezium . You have to divide the trapezium in two equal part of area . You have to draw a parallel line with the parallel line of the trapezium .

Consider the above trapezium . Given AB  ,CD , AD and BC . You have to find out AE and BF so that we can draw the parallel line joining E and F . Given trapeziums are as above . Where AB || CD and AB>CD .

Here , the area of ABFE = the area of CDEF = the area of trapezium ABCD / 2 .

The area of trapezium ABCD = ( AB + CD ) * h / 2 ------------------------ (1)

Here h is height and it is unknown .

We can draw a triangle combining the triangle AmD and BnC . The new triangle is pqr . We know its sides . pq = AB - CD , pr = AD and qr = BC . So we can calculate its area . Say it is triangleArea .
We know , area of triangle pqr , triangleArea  = pq * h / 2 .

                                              or , h= 2 * triangleArea / pq .

Putting the value of h in equation (1) , we may get the area of trapezium ABCD . Say it is trapeziumArea .

So , the area of ABFE = the area of CDEF = trapeziumArea / 2 .

Or , the area of CDEF = trapeziumArea / 2 = CD * n + triangle rst .
Or ,  CD * n +  b * n / 2 = A  .        --------------- (2)        Let , A =  trapeziumArea / 2 .

In the triangle pqr , rst and pqr are simillar triangle .

So , b / pq = n / h
So , b = n * pq / h

Putting the value of b in equation (2) ,

we get , CD * n + ( n * pq / h ) * n / 2 = A
       Or,  CD * n + pq * n * n / 2h = A
      Or , pq * n * n + 2h * CD * n - 2 * h * A = 0

So , n = ( -2h * CD + sqrt(( 2h * CD ) * ( 2h * CD) - 4 * pq * ( - 2A * h ) ) ) / ( 2 *pq )

In triangle AmD , n / h = DE / AD
                   So , DE = AD * n /h .
                   So , AE = AD - DE .

Similarly , in triangle MnC ,
                 n / h = CF / BC
            So , CF = BC * n / h
            So , BF = BC - CF .

And our answer is AE and BF .

My Solution