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

Join AD and consider the isosceles triangle ACD. Drop a perpendicular from C to AD and denote it as P

ReplyDeleteThe points A,C,E and P are concyclic since m(AEC) = m(APC) = 90 degrees

Join CB and let m(BAC)=X,m(BAD)=Y

=> m(DAC)=m(CDA)=X+Y

From angles in the same segment we have m(BDC)=X, m(DCB)=Y and m(CBA)=X+Y

Since ACEP is concylic, => m(EPC)=m(EAC)=m(BAC)=X and m(PCE)=m(PAE)=m(DAE)=Y

So the triangles APC and BEC are similar (AAA)

=> BE = AP.EC/PC ------------(1)

Similarly, triangle PEC and DBC are similar

=> BD = EP.CD/PC = PE.AC/PC -----------(2) (since CD=AC)

(1)+(2) => BE+BD = (AP.EC+PE.AC)/PC -----------(3)

Applying ptolmey's to the cyclic quadrilateral ACEP

=> AE.PC = AP.EC+AC.PE

=> AE = (AP.EC+PE.AC)/PC -----------(4)

From (3) and (4), BE+BD=AE Q.E.D