See complete Problem 119
Area of Triangles, Incenter, Excircle. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Tuesday, June 10, 2008
Elearn Geometry Problem 119
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name K point circle I meet BC, M meet AC, N meet AB
ReplyDeletename L point circle E meet BC
SILH = SICL - SICH
SILH = IK∙LC - SICH
SILH = r∙LC - SICH (1)
SHCF = SICF - SICH
SHCF = IM∙CF - SICH
SHCF = r∙CF - SICH
SHCF = r∙LC - SICH (2) ( CF = LC ,tg from C )
from (1) & (2)
SILH = SHCF (3)
in the same way
SILG = IBL - SIBG
SILG = r∙LB - SIBG (4)
SBGD = IBD - SIBG
SBGD = r∙BD - SIBG
SBGD = r∙LB - SIBG (5) (BD = LB, tg from B )
from (4) & (5)
SILG = SBGD (6)
add (3) & (6)
SILH + SILG = SHCF + SGBD
SIGH = SHCF + SBGD
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https://photos.app.goo.gl/qqgCBTGrUnMbmxsy6
DeleteLet s = semiperimeter of ABC.
ReplyDeleteAD = AF = s
Since area of ABC = rs and since
(area of AID) = (area of AIF) = (area of ABC)/2
(area of IDF) =(ADF) - [(ADI) + (AFI)]
=(ADF) - (ABC)
=(BCFD)
Or (IDF) = (BCFD). Now subtract (GHFD) from both sides to complete the proof.
Alternative solution to problem 119.
ReplyDeleteLet be p the semiperimeter of ABC and r the radius of its incircle. We have BD = p – c and CF = p – b.
So S1 = S(DBI) – S(BGI) = (p-c)r/2 – S(BGI) and S2 = S(CFI) – S(CHI) = (p-b)r/2 - S(CHI).
Besides S3 = S(BCI) – S(BGI) – S(CHI) =
= ar/2 – S(BGI) – S(CHF).
Thus
S1 + S2 = (p-c)r/2 + (p-b)/2 - S(BGI) - S(CHI) =
= (2p-b-c)r/2 - S(BGI) - S(CHI).
But 2p – b – c = a, so
S1 + S2 = ar/2 - S(BGI) - S(CHI) = S3.
Let r = inradius
ReplyDeleteEasily BD + CF = BC
So r.BD/2 + r.CF/2 = r.BC/2
Hence S(IBD) + S(ICF) = S(IBC)
So S1+S 2 = S3
Sumith Peiris
Moratuwa
Sri Lanka