Proposed Problem
Click the figure below to see the complete problem 433 about Quadrilateral, Triangle, Area, Proportion, Measurement, Similarity.
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Complete Problem 433
Level: High School, SAT Prep, College geometry
Sunday, April 4, 2010
Problem 433: Quadrilateral, Triangle, Area, Proportion, Measurement, Similarity
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draw BGH//CD, G on altitude from E, H on alt from A to CD
ReplyDeletename h1 altitude from B to CD, name h alt from E to CD
h2 alt from A to CD
name EG = x, AH = y
=>
h = h1 + x, h2 = h1 + y
have to prove
CF∙h1 + FD∙h2 = CD∙h ??? (1)
CF∙h1 + FD∙(h1+y) = CD∙(h1+x)
CF∙h1 + FD∙h1 + FD∙y = CD∙h1 + CD∙x
(CF + FD)∙h1 + FD∙y = CD∙h1 + CD∙x
FD∙y = CD∙x
x/y = FD/CD ??? (2)
BE/AB = x/y (3) from thales theorem
AE/BE = CF/FD is given (P431)
=>
AB/BE = CD/FD
=>
BE/AB = FD/CD (4)
from (3) and (4)
x/y = FD/CD ( 2 is proved) =>
(1) is proved
[ECA]/[BCA] = AE/AB = CF/CD = [ACF]/[ACD] = ([ECA]+[ACF])/([BCA]+[ACD]) = [ECFA]/[ABCD]
ReplyDeletesimilarly, EB/AB = [BFDE]/[ABCD]
add them up we get
AE/AB+EB/AB = [ECFA]/[ABCD]+[BFDE]/[ABCD] <=>
[ABCD] = [ECFA]+[BFDE] = [BFA]+[CDE] <=>
[ABCD]-[BFA] = [CDE] <=>
S1+S2 = S
This quadrilateral can be transformed to a triangle with other conditions unchanged.
ReplyDeleteExtending DC so that C is a vertex of tr(ADC) and the ratio CF:FD is unchanged,
the problem becomes like this
"The figure shows a triangle ADC with AE*DF=BE*CF. If S, S1, and S2 are the areas of triangles CED, BCF, and ADF, prove that S = S1 + S2."
let AE=1, BE=a, DF=b, CF=b/a, AB=d,
then, tr(AED)=tr(AFB)=tr(ADC)/d
.....
QED