## Tuesday, June 10, 2008

### Elearn Geometry Problem 119

See complete Problem 119
Area of Triangles, Incenter, Excircle. Level: High School, SAT Prep, College geometry

1. name K point circle I meet BC, M meet AC, N meet AB
name 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)

SILH + SILG = SHCF + SGBD

SIGH = SHCF + SBGD
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2. Let s = semiperimeter of ABC.
Since area of ABC = rs and since
(area of AID) = (area of AIF) = (area of ABC)/2
=(BCFD)
Or (IDF) = (BCFD). Now subtract (GHFD) from both sides to complete the proof.

3. Alternative solution to problem 119.
Let 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.

Easily 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