Problem 653: Diameter, chord, perpendicular

Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #13 (high school level) and lift up your geometry skills.

Click the figure below to see the complete problem 653.

Continue reading at:

gogeometry.com/ArchBooLem13.htm

## Tuesday, December 16, 2008

### Archimedes' Book of Lemmas, Proposition #13

Labels:
Archimedes,
book of lemmas,
chord,
circle,
diameter,
perpendicular

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The situation becomes clear, if you extend the lines AF and BG, so that they intersect the circle a second time.

ReplyDeleteLet then AF intersect the circle in P and let BG intersect the circle in Q. Then we have angles BPA=AQB=90° (Thales). As AP and BQ are parallel, QBA=PAB and ABP=BAQ as they are alternate angles; and as the sum of them is 180° (they are part of a cyclic quadrangle), each of them is 90°.

Thus APBQ is a rectangle. If we now draw the line through the center of the circle, which is perpendicular to CD, this will be a median of the rectangle (as a parallel to two sides which passes through the center). So, if we consider a reflection in this line, the circle and the rectangle are invariant and the segments DF and CG are mapped to each other. Thus they are equal. (by Thomas)

Let O is the center of circle and I is the projection of O over CD.

ReplyDeleteSince O is the midpoint of AB so I is the midpoint of FG and DC( F, I, G are the projection of A,O,B)

CG=CI-IG

DF=ID-IF

Since ID=IC and IF=IG so CG=DF

Peter Tran

AD = ABcos(BAD) & DF=ADcos(ADC) or DF=ABcos(ADC)cos(BAD) But /_ADC=/_ABC (angles in the same sector) & /_BCD=/_BAD for the same reason. Hence DF=ABcos(ABC)cos(BAD)=BCcos(BCD)=CG

ReplyDeleteAjit

Draw OH ⊥ CD. OH bisects CD

ReplyDeleteCH = HD

AF ∥ OH ∥ BG

Transversal AB is bisected at O

So transversal GF is bisected at H

GH = HF

Follows

CG = CH - GH = HD - HF = FD

produce BG to cut the circle at P,=>angle BAP=90

ReplyDeleteIf we observe AFPG, angles P,G,F are 90 degrees

=>angle FAP is 90 degrees,therefore AFPG is rectangle =>AF=PG

Now triangles AFD and PGC are similair and AF=PG

=>triangles AFD and PGC are congruent

=>CG=FD

http://img43.imageshack.us/img43/8577/u2mk.png

ReplyDeleteLet AF meet the circle at H.

ReplyDeleteDHBC is an isoceles trapezoid and HFGB is a rectangle.

Triangles BGC & HFD are therefore congruent ASA.

Hence DF = CG.

Sumith Peiris

Moratuwa

Sri Lanka