See complete Problem 232 at:

gogeometry.com/problem/p232_parallelogram_perpendicular_lines.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Sunday, January 25, 2009

### Elearn Geometry Problem 232: Parallelogram, Vertex, Perpendicular lines

Labels:
congruence,
distance,
parallelogram,
perpendicular,
trapezoid,
vertex

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Let O be the intersection of AC and BD, and O' the projection of O on the exterior line.

ReplyDeleteIn the triangle BB'D we have :

OO'=BB'/2=b/2

In the trapezoid AA'C'C we have:

OO'=(AA'+CC')/2=(a+c)/2

From these relations we obtain b=a+c

draw AB" perpendic to BB'

ReplyDeletetr ABB" and DCC' are similar ( ang B1 = C2 )

b-a /AB = c/CD => b-a = c => b = a+c

Let the perpendicular from C to BB' meet BB' at P.

ReplyDeleteby AAS congruence, triangle CPB congruent to DA'A.

hence PB = a.

now, PB'C'C is a rectangle...

so, PB' = c.

Hence b = BB' = BP + PB' = a + c .