Geometry Problem

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## Saturday, February 19, 2011

### Problem 579: Quadrilateral, Diagonals, Triangles, Areas

Labels:
area,
diagonal,
quadrilateral,
triangle

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Areas of triangles ABO & CBO are proportionals to bases: S1/S2 = AO/CO

ReplyDeleteAreas of triangles ADO & CDO are proportioanls to bases: S3/S4 = CO/AO

Then we've: S1/S2 = S4/S3 or S1·S3 = S2·S4.

Migue.

c.t.e.o has left the following solution of problem 579:

ReplyDeleteDraw BG, DH altitude on AC

AO∙BG = S1, CO∙BG = S2 => S1/S2 = AO/CO

AO∙DH = S3, CO∙DH = S4 => S4/S3 = AO/CO

=>

S1∙S3 = S2∙S4 (Q.E.D.)

S1=1/2.OA.OB.sin(AOB)

ReplyDeleteS2=1/2.OB.OC.sin(BOC)=1/2.OB.OC.sin(180-AOB)=1/2.OB.OC.sin(AOB)

S3=1/2.OC.OD.sin(COD)=1/2.OC.OD.sin(AOB)

S4=1/2.OD.OA.sin(AOD)=1/2.OD.OA.sin(180-AOB)=1/2.OD.OA.sin(AOB)

S1.S3=S2.S4=1/4.sin^2(AOB).OA.OB.OC.OD