Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the problem 714 details.
Thursday, January 5, 2012
Problem 714: Triangle, Parallel, Cevians, Midpoints, Median, Ceva Theorem
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Minor correction due to typo error:
ReplyDeleteApply Ceva's theorem in tri. ABC and cevians AE,CD and BG.
(DA/DB).(EB/EC).(GC/GA)=1 ----(1)
Since DE//AC >>>DA/DB=EC/EB
Replace this in (1) and simplify we get GC/GA=1 >> G is the midpoint of AC.
Using similar triangles with // lines we have
DH/AG=HE/GC=BH/BG >> HD=HE >> H is the midpoint of DE
Draw AM ⊥ BG(extended)and CN ⊥ BG.
ReplyDeleteDenote area by (...)
(ADC)=(AEC)(enclosed between parallels and having common base)
Remove common area (AFC) from either side:
(ADF) = (CEF)
Add common area (DBEF) to either side:
(BDC)=(BEA)
They have a common base BF
So AM = CN. Also AM ∥ CN
∆AMG ≡ ∆CNG (a.s.a)
∴AG = GC
AG being median bisects DE too (∵ DE ∥ AC)
To Pravin:( refer to your solution)
ReplyDeletearea of (BDC)= (BEA) is correct but these 2 triangles do not have common base.
Please explain.
Revised and corected Solution of Problem 714 (Thanks to Tran):
ReplyDeleteDraw AM ⊥ BG(extended)and CN ⊥ BG.
Denote area by (...)
(ADC)=(AEC)(enclosed between parallels and having common base)
Remove common area (AFC) from either side:
(ADF) = (CEF)
Add common area (DBEF) to either side:
(BDC)=(BEA) [as proved earlier]
Now,
AG/GC = (ABG)/(CBG) = (AFG)/(CFG)
= Diff of Nrs / Diff of Drs
= (ABF)/(CBF) --->(1)
By similar triangles AG/HE = FG/HF = GC/DH
∴ AG/GC = HE/DH --->(2)
Next HE/DH = (BHE)/(BHD) = (HFE)/(HFD)
=Sum of Nrs./Sum of Drs.= (BFE)/(BFD)--->(3)
From (1), (2), (3)
AG/GC = HE/DH = (ABF)/(CBF) = (BFE)/(BFD)
= sum of Nrs./Sum of Drs.= (ABE)/(BCD) =1
Hence AG = GC and DH = HE