Geometry Problem. Post your solution in the comment box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Friday, June 28, 2019

### Geometry Problem 1440 Intersecting Circles, Perpendicular Bisector, Collinear Points

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If AC extended meets circle Q at E’, Tr.s OAC & E’CD are iscoceles and similar

ReplyDeleteThen E’M is perpendicular to CD. But EM is also perpendicular to CD

Hence E and E” coincide and the result follows

Sumith Peiris

Moratuwa

Sri Lanka

Bonus : EM intersects circle (Q) at N.

DeleteThen, by the same method, N, C, B are collinear.

Extend OQ to meet cirlce Q at F and form the diameter OF.

ReplyDeleteLet m(EDF)=x => ECD and OAC are two similar isosceles triangles with angles (2x,90-x,90-x)

=> AC/OC=CD/EC

=>AC.EC=OC.CD

Since AE and OD are two chords of Q => A,C,E must be collinear

If:

ReplyDeleteOC=r , OD=a => CM=MD=(a-r)/2 , OM=(a+r)/2

Define:

ME=p

So:

tg(MEC)=(a-r)/2p => 90-arctg((a-r)/2p)=MCE

tg(MEO)=(a+r)/2p => 90-arctg((a+r)/2p)=MOE

We know:

1.) OAC=OCA and AOC=alpha => OAC=90-alpha/2

2.) Points A O E give a Tri., so AOE+OEA+OAE=180

So:

MEO-MEC+AOC+OAC+MOE=180 => alpha=2*arctg(a-r)/2p=2*MEO

_____

90-alpha/2=MCE => OCA=MCE

So AC and CE are on the line.