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

Labels:
Ceva's theorem,
cevian,
concurrent,
median,
midpoint,
parallel,
triangle

Subscribe to:
Post Comments (Atom)

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