Geometry Problem

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## Thursday, April 28, 2011

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## Thursday, April 28, 2011

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Geometry Problem 595: Triangle, Equal Angles, Isogonal Conjugate, Similarity

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angle,
angle bisector,
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Law of sinus:

ReplyDelete(1) Tr. ABC c/a = sin(C)/sin(A)

(2) Tr. ABD d1/sin(alpha) = BD/sin(A)

(3) Tr. CBE e2/sin(alpha) = BE/sin(C)

(4) Tr. ABE e1/sin(B-alpha) = BE/sin(A)

(5) Tr. CBD d2/sin(B-alpha) = BD/sin(C)

From (1)&(2)&(3): d1/e2 = c·BD/(a·BE) (6)

From (1)&(4)&(5): e1/d2 = c·BE/(a·BD) (7)

From (6)&(7): d1·e1/(d2·e2)= c·c/(a·a)

MIGUE.

Let the circle through B, D, E

ReplyDeleteintersect AB at R and BC at S

AD.AE = AR. AB and CE.CD = CS.CB

Arcs DR, ES are of equal length

(since they subtend equal angles at B)

So RS ∥DE or RS ∥AC

AR/AB = CS/CB

(or) AR/CS = AB/CB

Hence

(d1.e1)/(d2.e2)

= (AD.AE)/(CE.CD)

= (AR.AB)/(CS.CB)

= (AR/CS).(AB/ CB)

= (AB/CB).(AB/ CB)

= c^2/a^2