Question 1043167
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ABCD is a rectangle with M the midpoint of BC. AC intersects MD at N. Find area of triangle NCD and area of quadrilateral ABMN.
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See the Figure on the right. 

Look in it attentively and identify all given lines and points. 

In addition to the given lines, I drew  the line BK with the 

point K as the middle point of the side AD. 

Let L be the intersection point of BK and AC.

Then it is clear from symmetry that |AL| = |CN| (congruent, have equal lengths).

Also from symmetry, it is clear that |AN| = |LC|.

   (One could find more complicated arguments, but it is enough for me now).

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{{{drawing( 420, 220,  -0.5, 4.5, -0.5, 2.5, 
            line( 0,  0, 4,  0), 
            line( 0,  0, 0,  2),
            line( 4,  0, 4,  2),
            line( 0,  2, 4,  2),

            locate(-0.1,  0.0,  A),
            locate( 4.0,  0.0,  B),
            locate(-0.1,  2.2,  D),
            locate( 4.0,  2.2,  C),

            line( 0,  2, 4,  1),
      green(line( 0,  0, 4,  2)),
            locate( 4.05,  1.1,  M),
            locate( 2.55,  1.6,  N),

       blue(line( 0, 1, 4,  0)),
            locate( -0.15,  1.1,  K),
            locate( 1.33,  0.65, L)

)}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<B>Figure</B>. 

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</TABLE>It implies that  |AL| = |LN| = |NC| = {{{1/3)}}}|AC|.   (Actually, it is well known property for this situation).
               (see the lesson <A HREF=https://www.algebra.com/algebra/homework/word/geometry/Solved-problems-on-Parallel-lines-cutting-off-congruent-segments-in-transverse-lines.lesson>Solved problems on Parallel lines cutting off congruent segments in transverse lines></A> in this site).


If so, then the height of the triangle DNC is exactly one third of the side |BC|.

Thus the area of the triangle DNC is one sixth ({{{(1/2)*(1/3)}}}) of the area of the rectangle.

Regarding the quadrilateral ABMN, one can show that its area is {{{(1/2) - (1/12)}}} = {{{5/12}}} of the area of the rectangle.
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