## Friday, August 29, 2008

### Elearn Geometry Problem 166

See complete Problem 166 at:
www.gogeometry.com/problem/p166_parallelogram_triangle_area.htm

Parallelogram, Diagonal, Triangles, Areas. Level: High School, SAT Prep, College geometry

1. draw diagonal AC which meets BD at G and BE at H.also join GE.
proof:we know that in a parallelogram diagonals bisect each other.so AG=GC and BG=GD.
also median of a triangle divides it into two triangles of equal area.
[tri BGE]=[tri GED]
[tri BGH]+[tri GHE]=[tri GEF]+[tri BFD]-----( 1 )
[tri AEG]= [tri GEC]
[tri AFG]+[tri GEF]=[tri GHE]+[tri HEC]------(2)
On adding 1 and 2 equations
[tri BGH]+[tri AFG]=[tri EFD]+[tri HEC]
On adding [tri AGB] on either side
[tri ABF]+[tri BGH]=[tri AGB]+[tri EFD]+[tri BHC]
but diagonals of a parallelogram divide it into four triangles of equal area.
[tri AGB]=[tri BGC] substitute in the above equation
and after simplifications we get the required result. S=S1+S2

2. Let A be (0,0), B;(b,h), C:(b+d,h) and D:(d,0) and let E be (p,q) Now we can determine the F as
(dhp/(ph-qb+qd),dhq/(ph-qb+qd)). Now Tr. AFD =d*dhq/2(ph-qb+qd)= hqd^2/(ph-qb+qd) -----(1)
While, Tr. BFE =(1/2)(bdhq/(ph-qb+qd) -bq+dhpq/(ph-qb+qd)-pdh^2/(ph-qb+qd) +ph -pdhq/(ph-qb+qd))and Tr. ECD =(dh+bq-hp)/2
Now Tr. BFE + Tr. ECD = hqd^2/(ph-qb+qd) upon addition and simplification. In other words, Tr. AFD = Tr. BFE + Tr. ECD because of (1)
In parallelogram ABCD, Tr. ABD = Tr. BDC or S + Tr. AFD = S1 + S2 + Tr. BFE + Tr. ECD or S = S1 + S2 since Tr. AFD = Tr. BFE + Tr. ECD
Ajit: ajitathle@gmail.com

3. Es mucho más fácil: aplicación directa del teorema de la alfombra.
Y por otro: S1+S2+[AFD]=0'5*[ABCD]
Por tanto, S=S1+S2

4. Solution to problem 166.
Let be h1 and h2 the distances from E to lines BC and AD, respectively, and h3 the distance from F to line AD. We have BC = AD.
Then S1 = BC.h1/2 = AD.h1/2,