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AF=AE=p-a=BM=CHBF=BD=p-b=AM=CGCD=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)sinBsinA=a/(2R);sinB=b/(2R);sinC=c/(2R)with a "little" algebraS_i=S_e
In the solution of pr. 86 posted by Anonymous, could anyone explain me why is it true thatAF = BM = CH = p - a ?Thanks for the help.Nilton Lapa
To Nilton, Problem 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.
of the above solution, why it can be said that the triangle DEF = triangle MGH??
In the solution of pr. 86 posted by Anonymous, could anyone please explain me how to do this ''little'' algebra?
Share your solution or comment below! Your input is valuable and may be shared with the community.
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