Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Tuesday, November 4, 2014
Geometry Problem 1055: Triangle, Altitude, Perpendicular, Area, Circumradius, Circle
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Let S be the area. Let BC=a, AC=b, AB=c.
ReplyDeleteSince ΔABC~ΔDBE, thus DE/AC=BD/AB.
d / b = (c sinA sinC) / c
d = b sinA sinC = b (a/2R) (c/2R)
Hence, S = abc/4R = Rd.
D,F,E,B are concyclic and they lie on the circle on BF as diameter.
ReplyDeleteIn this circle d = chord EF = (diameter BF) x sin B
(i.e.) d = h Sin B where h denotes height BF.
In ΔABC we have
2Δ= h.b = h. 2R sin B = 2R .h Sin B = 2R.d
Hence Δ = R.d
R=BC/2*BA/BF
ReplyDeleted=AC*BD/BA
Rd=BC*AC*BD/(2BF)
Due to similar triangles BC*BD=BF^2
Therefore, Rd=AC*BF/2=[ABC]
Considering usual triangle notations, it can be observed that BEFD is cyclic
ReplyDelete=> m(BED)=m(BFD)=C
Hence Tr. BEF similar to BCA
=> BE=ad/b
Mark O as circumcenter and G as mid-point of BC => BG=a/2
Observe that Tr.BOG is similar to BFE
=>BF=BO.BE/BG
=>BF=R.(ad/b)/(a/2)
=>BF.b/2=R.d
Q.E.D