## Sunday, August 31, 2008

### Elearn Geometry Problem 169

See complete Problem 169 at:
www.gogeometry.com/problem/p169_parallelogram_triangle_pentagon_area.htm

Parallelogram, Interior and Exterior Points, Diagonals, Pentagon, Triangles, Areas. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

1. G and H are 2 points on BC and BA respectively such that BA//EG and BC//HF,
[ABE]+[CDE]=[ABG]+[CDG]=[ABG]+[CAG]=[AHC]+[CHB]=[AHD]+[CHB]=[AFD]+[CFB]

2. Solution to problem 169.
Draw through E a line perpendicular to AB and CD, intersecting AB at F and DC at G.
Let be FG = h.
We have S(ABE) + S(EDC) = AB.FE/2 + CD.EG/2 = AB.h/2 (since AB = CD).
But AB.h/2 = P/2 (where P is the area of the parallelogram ABCD).
Let be X and Y the areas of the no shaded regions with vertex B and C respectively.
Then S1 + X + S2 + S3 + Y + S4 = P/2 (1).
Let be h5 and h6 the distances of point F to lines AD and BD respectively.
We have S(ADF) + S(BCF) = AD.h5/2 + BC.h6/2 = AD(h5+h6)/2 = P/2,
so S5 + X + S6 + Y = P/2 (2).
From (1) and (2) we get
S5 + X + S6 + Y = S1 + X + S2 + S3 + Y + S4,
or S5 + S6 = S1 + S2 + S3 + S4.