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

Labels:
area,
proportions,
quadrilateral,
similarity,
triangle

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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