Geometry Problem

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

Click the figure below to see the complete problem 763.

## Friday, June 8, 2012

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

Labels:
congruence,
diagonal,
parallel,
parallelogram

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AG/GC = AE/DC = AE/AB = CF/BC = CH/AH

ReplyDelete1 + AG/GC = 1 + CH/AH

AC/GC = AC/AH

GC = AH

GC - GH = AH - GH

CH = AG

Since EF//AC, so let BE/EA = BF/FC = k.

ReplyDeleteThen, FH/HD = EG/GD = 1/(k+1).

Let AD=BC=a, AB=CD=b, then

S(ADE) = 1/2*a*b*sin A*1/(k+1)

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

then S(ADE)=S(CDF) (notice that A=C).

Therefore,

S(ADG)

= S(ADE)*DG/DE

= S(ADE)*(k+1)/(k+2)

= S(CDF)*(k+1)/(k+2)

= S(CDF)*DH/DF

= S(CDH)

Since ADG and CDH have the same height,

thus, AG = CH.

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

ReplyDeleteFrom EF//AC,

so BE/EA = BF/FC and EG/GD = FH/HD.

AG/CH

= S(ADG)/S(CDH)

= S(ADE)/S(CDF)

= S(ADB)/S(CDB)

= 1

Hence, AG=CH.

Using the vector approach to solve the problem.

ReplyDeleteDenote 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.

Let BD intersect EF, AC, at L and O respectively.

ReplyDeleteIn ∆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

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

ReplyDelete