See complete Problem 86

Intouch and Extouch Triangles, Area. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

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## Monday, May 19, 2008

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Geometry Problem 86

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

See complete Problem 86

Intouch and Extouch Triangles, Area. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

AF=AE=p-a=BM=CH

ReplyDeleteBF=BD=p-b=AM=CG

CD=CE=p-c=BG=AH

[DEF]=[ABC]-0.5(p-a)²sinA-0.5(p-b)²sinB

-0.5(p-c)²sinC

[MGH]=[ABC]-0.5(p-b)(p-c)sinA-0.5(p-a)(p-b)sinC

-0.5(p-a)(p-c)sinB

sinA=a/(2R);sinB=b/(2R);sinC=c/(2R)

with a "little" algebra

S_i=S_e

In the solution of pr. 86 posted by Anonymous, could anyone explain me why is it true that

ReplyDeleteAF = BM = CH = p - a ?

Thanks for the help.

Nilton Lapa

To Nilton,

DeleteProblem 86, See

Semiperimeter s, Incircle and Excircles" or

Semiperimeter Index" or

use "two tangent segments to a circle from an external point are congruent."

To Antonio: thanks for the hints. My doubt is solved now.

ReplyDeleteof the above solution, why it can be said that the triangle DEF = triangle MGH??

ReplyDeleteIn the solution of pr. 86 posted by Anonymous, could anyone please explain me how to do this ''little'' algebra?

ReplyDelete