Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
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Saturday, September 24, 2016
Geometry Problem 1265: Triangle, Incircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Angle Bisector
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https://goo.gl/photos/mniwJNrj8t6jnXvG7
ReplyDeleteLet ED meet AC at P.
Apply Menelaus’s theorem in secant EDP of triangle ABC
PC/PA x EA/EB x DB/DC= 1
Replace EB=DB in above expression and simplify it . we get PC/PA=DC/EA= FC/FA
So (ACFP)= -1
Apollonius circle with diameter FP will pass through G
So GF is an angle bisector of angle AGC
A different solution
Deletehttps://photos.app.goo.gl/nkRnMwi6ABmJDBmh6
Let ED meet AC at P.
Let CM // AE ( M on DP; see sketch)
∠(CMD)=∠(BED) and ∠(CDM)=∠(EDE)
but triangle EBD is isosceles so triangle CDM is also isosceles
so PC/PA= CM/AE= CD/AE= FC/FA
So (ACFP)= -1
Apollonius circle with diameter FP will pass through G
So GF is an angle bisector of angle AGC
Problem 1265
ReplyDeleteIs BE=BD then <BED=<BDE or <AEG=<CDG.But <FOC=<FOD/2=<FEG and <CFO=90=<FGE so
Triangle EGF is similar the triangleOFC so EG/OF=GF/FC similar GD/OF=GF/AF with division by states we have EG/GD=AF/FC.So triangleAEG is similar triangleGDC then <EGA=<DGC.
Therefore <AGF=<FGC (as equal angles supplements).
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE
Really excellent work Apostolis
ReplyDeleteThank you for your nice words Sumith.
DeleteTrigonometry solution
ReplyDeleteEG = 2.AE.sin(A/2).sin(C/2)
GD = 2.CD.sin(C/2).sin(A/2)
So EG/GD = AE/CD
Hence Tr.s AEG and CDG are similar and the result follows
Sumith Peiris
Moratuwa
Sri Lanka
Geometry Solution
ReplyDeleteLet M be the midpoint of EF and let EF =m, FD =n, EG = p and GD = q
Tr.s EGM and DCM are similar (isosceles and both having <C), so
(m/2)/p = (s-c)/n ….(1)
where 2s = a+b+c with the usual notation.
Similarly we can show that
(n/2)/q = (s-a)/m…(2)
(2)/(1) gives, p/q = (s-a)/(s-c)
Hence Tr.s AEG and CDG are similar and the result follows.
Sumith Peiris
Moratuwa
Sri Lanka
To prove GF is angle bisector of AngleAGC, it would suffice to prove that the triangles AEG and CDG are similar.
ReplyDeleteLet A,B,C be the angles of the triangle ABC
Join EF and form the isosceles triangle AEF. Similarly join FD and form the isosceles triangle CDF.
In triangle AEF the angles are A, 90-A/2, 90-A/2
=> Angle GDF = 90-A/2 ( per Ext Segment theorm) => AngleGFD = A/2
Similarly AngleGFE = C/2
Join OE and consider the triangle OAE. The Angles in the triangle are 90,A/2 & 90-A/2 which is similar to the triangle GFD
Hence AE.GD = OE.GF -------------(1)
Similarly join OD and form the triangle ADC whose angles are 90,C/2 & 90-C/2, which is similar to the triangle GFE
Hence EG.DC = OD.GF
=> EG.DC = OE.GF (Since OE=OD) ---------(2)
FROM (1) & (2) AE.GD = EG.DC => Triangles AEG and CDG are similar
Therefore AG/AE = CG/CD
=> AG/AF = CG/CF ----------- (3)
Hence (3) is the result of Angle bisector theorem and GF is the bisector of Angle AGC
Its true for any cevian triangle DEF.
ReplyDeleteYes, it is true because the point of the projection of F over DE always stay in the Apollonius circle with diameter FP
Delete