See complete Problem 232 at:
gogeometry.com/problem/p233_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 233: Parallelogram, Exterior line, Perpendicular lines
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Let be O the intersection of AC and BD, and O' the projection of O on the exterior line. Let denote by o the lenght of OO'.
ReplyDeleteIn the trapezoid BB'D'D :
o=(b+d)/2
In the trapezoid AA'C'C :
o=(a+c)/2.
From this relations we have that :
a+c=b+d
like P234, 235
ReplyDeletedraw AB", DC" // A'C'
b-a /AB = c-d/CD
b-a = c-d => b+d = c+a
Draw D'D''DD''' parallel to A'B'D'C' thro' D.
ReplyDeletewhere D',D'',D''' lie on AA', BB', CC' resp.
now, D'A' = D''B' = DD' = D'''C'.
so, it requires us to prove that AD' + CD''' = BD'' which comes from problem 232.
Proof follows easily from my proof of Problem 232
ReplyDeleteExtend A`A to form a rectangle with C (A'ECC'). Let BB' meet EC at F => BF=b-c
DeleteJoin AC to form another rectangle A'ACC'. Let DD' meet AC at G => DG=a-d
BFC and DGA are congruent => BF=DG and hence a+c=b+d
Draw a line at A"B"C" pass through D // A'B'D'C'
ReplyDeleteFrom problem 232, AA"+CC"=BB" & d=A"A=B"B=C"C
a+c=AA"+A"A'+CC"+C"C
=AA"+CC"+A"A"+C"C'
=BB"+d+d
=BB"+B"B'+d
=b+d