Problem 652: Diameter, chords, perpendicular, tangents

Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #12 (high school level) and lift up your geometry skills.

Click the figure below to see the complete problem 652

Continue reading at:

gogeometry.com/ArchBooLem12.htm

## Tuesday, December 16, 2008

### Archimedes' Book of Lemmas, Proposition #12

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

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http://img193.imageshack.us/img193/4671/problem652.png

ReplyDeleteConnect OC , DE

AD cut BF at L ( see picture)

Quadrilateral LDFE is cyclic with LF as diameter ( Angles D=E = 90)

Note that angle(DAE)= angle (CDE)

So angle (DLE)= angle (DCO) ( angles complement to congruence angles)

OC is perpendicular bisector of DE . On OC only center of circle LDFE have property (DLE)= ½ (DCE)

So C is the center of circle LDFE and diameter LF pass through center C

In Triangle ABL , F is the orthocenter so line LCF will perpendicular to AB

Peter Tran

Hi Peter!

DeleteYou mean "AD cut BE at L ( see picture)" ?

Yes

DeleteNice solution!

ReplyDeletePeter Tran's solution suggests the following (perhaps easier) problem:

ReplyDelete"Let D, E be any two points on the semicircle on AB as diameter. Let AE, BD intersect at F; AD, BE intersect at L. Prove LF is perpendicular to AB."

Pravin

Peter Tran's solution gives me this idea:

ReplyDeleteTo construct the tangents we draw the circle with diameter CO

This cercle intersects the given semi circle at D and E

he passes trough O midpoint of AB and the altitude feet E,D, he is the nine point circle of triangle ABL

let be M the second intersection with AB

M is an altitude foot

CM perpendicular to AB (CO diameter)

hence CM is an altitude and passes trough orthocenter F and M=G

FC perpendicular to AB

Let AD, BE meet at M. MF is obviously perpendicular to AB

ReplyDeleteLet DC meet MF at S and EC meet MF at T

< SDM = < SMD = 90 - A so DS = SM ....(1)

Similarly MT = TE .....(2)

DC = CE......(3)

(1) +(2) -(3) yields DS + MT - DC = MS + ET - CE which in turn yields SC+CT+TS = 0 which can only happen if C,T,S coincide

So C must lie on MF which completes our proof

Sumith Peiris

Moratuwa

Sri Lanka

Hello Sumith.

DeleteIn your proof, the sentence "DS + MT - DC = MS + ET - CE which in turn yields SC+CT+TS = 0 which can only happen if C,T,S coincide" seems to infer that SC, CT and TS should all be positive or equal to 0.

How can we sure about that ?

Best regards.

Link doesnt work

ReplyDelete