Geometry Problem
Click the figure below to see the complete problem 595.
Thursday, April 28, 2011
Geometry Problem 595: Triangle, Equal Angles, Isogonal Conjugate, Similarity
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Geometry Problem
Click the figure below to see the complete problem 595.
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