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
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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 .
Complete Rectangle A'B'BX
ReplyDeleteFrom< ABX = < CDC' so Tr.s ABX and CC'D are congruent ASA
So AX = c and hence b = A'X = a + c
Sumith Peiris
Moratuwa
Sri Lanka