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

Labels:
altitude,
area,
circle,
circumradius,
perpendicular,
triangle,
tutorial

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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