See complete Problem 154
Triangle, Inradius, Circumradius, Chord. Level: High School, SAT Prep, College geometry
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Sunday, August 3, 2008
Elearn Geometry Problem 154
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1)in triangle DIC
ReplyDeleteang(DIC)=ang(BDC)=ang(BAC)=A;inscribed angles with endpoints B and C
the lines BI and CI are bissectors ang(DIC)=(B+C)/2;ang(ICD)=A+B+C-A-(B+C)/2=(B+C)/2
triangle DIC is isoscele DI=DC
2)with the law of sines
in triangle IBC: BC sin(C/2)=IB sin((B+C)/2)
in triangle IDC: IC sin((B+C)/2)=ID sinA
in triangle ABC: BC= 2RsinA and r= ICsin(C/2)
therefore 2Rr=IB.ID
other questions related to this configuration
ReplyDeletea) prove that D is the circumcenter of triangle AIC
b)lines AI,CI intersect the circumcercle of ABC at E and F.
Prove that I is the orthocenter of DEF
(1) Since BD bisects angle B, D is the midpoint of arc ADC. Therefore B/2 = CAD = ACD, and ICD = C/2 + ACD = C/2 + B/2. On the other hand, DIC = IBC + BCI = B/2 + C/2. Hence DIC is isosceles and DI = CD.
ReplyDeletehttp://ahmetelmas.wordpress.com/2010/05/15/geo-geo/
ReplyDeletehttp://img208.imageshack.us/img208/1433/respuesta154.png
ReplyDelete< ICA = C/2, < ACD = B/2 and < BDC = A
ReplyDeleteSo < DIC = B/2 + C/2, so ID = CD
If CO meets the circumcircle at X, < BXD = B/2 and from similar triangles,
CX/CD = BI/r from which our result follows since CX = 2R
Sumith Peiris
Moratuwa
Sri Lanka