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.

## Sunday, August 31, 2008

### Elearn Geometry Problem 169

Labels:
area,
diagonal,
exterior point,
interior point,
parallelogram,
pentagon,
triangle

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G and H are 2 points on BC and BA respectively such that BA//EG and BC//HF,

ReplyDelete[ABE]+[CDE]=[ABG]+[CDG]=[ABG]+[CAG]=[AHC]+[CHB]=[AHD]+[CHB]=[AFD]+[CFB]

Solution to problem 169.

ReplyDeleteDraw 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.