Proposed Problem

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

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Complete Problem 431

Level: High School, SAT Prep, College geometry

## Thursday, April 1, 2010

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

Labels:
diagonal,
midpoint,
quadrilateral,
similarity,
transversal

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draw AKL//GH, (K on BD, L on CD)

ReplyDeletedraw 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

bc = ad

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

ReplyDelete1. 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

Problem 431

ReplyDeleteFetch 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

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

ReplyDeleteextended 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