## Wednesday, September 30, 2009

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

Proposed Problem
Click the figure below to see the complete problem 362 about Circle, Chord, Perpendicular, Equal chords. See more:
Complete Problem 362
Level: High School, SAT Prep, College geometry

1. join C with B, draw CFL ( F on AB, L on circle) that
FE = 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

2. 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
Hence Tr.s CDB & CFB are congruent and so BD=BF
But AE=EF hence AE = EB + BD

Sumith Peiris
Moratuwa
Sri Lanka

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