## Friday, June 8, 2012

### Problem 763: Parallelogram, Diagonal, Parallel, Congruence

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 763.

1. AG/GC = AE/DC = AE/AB = CF/BC = CH/AH
1 + AG/GC = 1 + CH/AH
AC/GC = AC/AH
GC = AH
GC - GH = AH - GH
CH = AG

2. Since EF//AC, so let BE/EA = BF/FC = k.
Then, FH/HD = EG/GD = 1/(k+1).

S(CDF) = 1/2*a*b*sin C*1/(k+1)

Therefore,
= S(CDF)*(k+1)/(k+2)
= S(CDF)*DH/DF
= S(CDH)

Since ADG and CDH have the same height,
thus, AG = CH.

3. Denote the area of triangle PQR by S(PQR).

From EF//AC,
so BE/EA = BF/FC and EG/GD = FH/HD.

AG/CH
= 1

Hence, AG=CH.

4. Using the vector approach to solve the problem.
Denote vector DA = a, vector DC = c.
Let AE:EB=k:1-k, then BF:FC=k:1-k (as triangle BEF ~ triangle BAC)
By considering that AG is along AC , AG = rAC = -ra + rc
By considering that AG divides DE into s:1-s, AG = -sa + k(1-s)c
Hence r = s = k/(1+k)
=> AG = (k/(1+k))(c-a)
Symmetrically, CH = -(k/(1+k))(c-a)
Q.E.D.

5. Let BD intersect EF, AC, at L and O respectively.
In ∆ABC,
O is the midpoint of AC and EF ∥ AC imply L is the midpoint of EF.
In ∆DEF,
L is the midpoint of EF and GH ∥ EF imply O is the midpoint of GH.
Hence
AG = OA - OG = OC - OH = CH

6. http://www.mathematica.gr/forum/viewtopic.php?f=20&t=27378&p=133967

7. Let BD meet EF at N and AC at M.

Since EF // AC and AM = MC, EN = NF which implies that GM = MH and so since AM = MC, AG = HC

Sumith Peiris
Moratuwa
Sri Lanka

8. https://photos.app.goo.gl/zzGJTYLenTIJ0U7E2
let DB cut AC and EF at O and I
We have O and I are midpoints of AC and EF
In triangle DEF since I is the midpoint of EF and GH//EF
So O is the midpoint of GH => AG=HC