Let P be a point on side AB of triangle ABC.
Construct a line through P that divides the
triangle into two polygons of equal area.
case1: PA < PB,
D will be the point on BC such that PD
divides ABC into two polygons of equal area.
Area of PDB = Area of ABC/2
1/2 * PB * DB * sinB = 1/2 *1/2 * AB * BC
2*PB*DB=AB*BC
2*BP/BA = BC/BD
So, extend the line BA and find the point X such
that BX = 2*BP
Join XC
Draw the line parallel to XC through P, it will
intersect BC in D.
PD is the required line.
case2: PA > PB,
D will be the point on AC such that PD
divides ABC into two polygons of equal area.
same repeats
case3: PA = PB
PA is the required line.
Edited on July 7, 2010, 4:21 pm
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Posted by Praneeth
on 2010-07-07 16:18:20 |
Solution