Geometry Problem

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## Saturday, February 19, 2011

### Problem 580: Triangles, Equal Angle, Transversal, Product of Sides, Areas

Labels:
area,
product,
transversal,
triangle

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draw DH1, AH altitudes on BE, BC

ReplyDeleteBE∙ DH1 = S1, BC∙AH = S => S1/S = (BE∙DH1)/(BC∙AH)

from ▲DBH1, ▲BAH => DH1/AH = BD/AB substitute above

S1/S = (BE∙BD)/(BC∙AB)

(I have sent yesterday solution of 579)

Join point E & point A. Then area of triangle BED (S1) is proportional to area of triangle BEA (S2), and the ratio is:

ReplyDeleteS1/S2 = BD/BA. (1)

Now area of triangle BAC (S) is proportional to area of triangle BEA (S2): S/S2 = BE/BC. (2)

We multiply (1) & (2) to obtain the result:

(S1/S2)·(S/S2) = S1/S = BD·BA/(BE·BC).

MIGUE.

Sorry:

ReplyDeleteS/S2=BC/BE

(S1/S2)•(S2/S)=BD•BE/(BA•BC)

S=1/2.BA.BC.sinB

ReplyDeleteS1=1/2.BD.BE.sinB