## Thursday, April 1, 2010

### Problem 431: Quadrilateral, Midpoints of Diagonals, Transversal

Proposed Problem
Click the figure below to see the complete problem 431 about Quadrilateral, Midpoints of Diagonals, Transversal, Similarity.

Complete Problem 431

Level: High School, SAT Prep, College geometry

1. draw AKL//GH, (K on BD, L on CD)
draw QPC//GH, (Q on AB, P on BD)
=>
GE middle line for ACQ ( E midpoint and QC//GH)
=> G midpoint
=> AG = GQ = a
in the same way
=> CH = HL = c
from thales theorem in AQPK
GQ/GA = PF/FK => a/a = PF/FK
=>
PF = FK (1)
BF = FD (2) ( F midpoint of BD )

from thales theorem i ▲ABK
BG/GA = BF/FK
b/a = BF/FK (3)

from thales theorem in ▲DCP
DH/HC = DF/FP
d/c = DF/FP (4)

from (1) and (2)
BF/FK = DF/FP (5)

from (3) and (4) and (5)

b/a = d/c

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2. Drawing CN//AB (N on line GH) and DM//AB (M on line GH)

1. Note that Triangle AGE congruent to Triangle CNE ( case ASA) so CN=AG=a
Triangle BFG congruent to Triangle DFM ( case ASA) so DM=BG=b
2. Triangle CHN similar to triangle DHM ( case AA) so CH/HD=CN/DM or c/d=a/b
3. b*c=a*d

Peter Tran

3. Problem 431
Fetch from point B parallel to the GΗ intersecting straight AC , DC in points K,L respectively.
AG/GB=AE/EK,DH/HL=DF/FB so DH=HL. But CH/DH=CH/HL=CE/EK=AE/EK. Then
AG/BG=CH/DH or AG.DH=BG.CH.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

4. Draw a parallel thro’ A to meet EG at X. Draw a parallel thro’ D to meet FH
extended at Y

Triangle pairs {BFG, DFY} and {AEX, CEH} are congruent so AX = c and DY = b.

Further triangles AXB and DHY are easily seen to be similar so c/d = a/b and the result follows

Sumith Peiris
Moratuwa
Sri Lanka