Proposed Problem

Click the figure below to see the complete problem 362 about Circle, Chord, Perpendicular, Equal chords.

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Complete Problem 362

Level: High School, SAT Prep, College geometry

## Wednesday, September 30, 2009

### Problem 362. Circle, Chord, Perpendicular, Equal chords

Labels:
chord,
circle,
equal chords,
perpendicular

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join C with B, draw CFL ( F on AB, L on circle) that

ReplyDeleteFE = EB ( get ang CBE = ang CFB )

now have to prove AF = DB

ang CBE = ang ALF (at the same harc AC )

=> ang ALF = ang AFL ( AFL = CFE = CBE )

=> AL = AF

now have to prove AL = BD

ang FLD = CBE ( on equal harcs AC=CBD)

=>ang FLD = ang AFL

=> AB // LD, => AL = BD, or AF = BD

Extend AB to F such that AC = CF. Then C is the circumcentre of Tr. ADF. Hence < DCF = 2.< DAF and since< DAC =< DCB it follows that BC bisects< DCF

ReplyDeleteHence Tr.s CDB & CFB are congruent and so BD=BF

But AE=EF hence AE = EB + BD

Sumith Peiris

Moratuwa

Sri Lanka