document.write( "Question 1043167: ABCD is a rectangle with M the midpoint of BC. AC intersects MD at N. Find area of triangle NCD: area of ABMN \n" ); document.write( "

Algebra.Com's Answer #658358 by ikleyn(52781) $\"\"$ $\"About$

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\n" ); document.write( "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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document.write( "See the Figure on the right. \r\n" );
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document.write( "Look in it attentively and identify all given lines and points. \r\n" );
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document.write( "In addition to the given lines, I drew  the line BK with the \r\n" );
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document.write( "point K as the middle point of the side AD. \r\n" );
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document.write( "Let L be the intersection point of BK and AC.\r\n" );
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document.write( "Then it is clear from symmetry that |AL| = |CN| (congruent, have equal lengths).\r\n" );
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document.write( "Also from symmetry, it is clear that |AN| = |LC|.\r\n" );
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document.write( "   (One could find more complicated arguments, but it is enough for me now).\r\n" );
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document.write( "                    Figure. \r\n" );
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It implies that  |AL| = |LN| = |NC| = |AC|.   (Actually, it is well known property for this situation).\r\n" );
document.write( "               (see the lesson Solved problems on Parallel lines cutting off congruent segments in transverse lines> in this site).\r\n" );
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document.write( "If so, then the height of the triangle DNC is exactly one third of the side |BC|.\r\n" );
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document.write( "Thus the area of the triangle DNC is one sixth () of the area of the rectangle.\r\n" );
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document.write( "Regarding the quadrilateral ABMN, one can show that its area is  =  of the area of the rectangle.\r\n" );
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