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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## Tuesday, June 10, 2008

###
Elearn Geometry Problem 119

Online Geometry theorems, problems, solutions, and related topics.

See complete Problem 119

Area of Triangles, Incenter, Excircle. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

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

------------------------------------------

Let 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