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

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

Details: Click on the figure below.

## Sunday, January 15, 2017

### Geometry Problem 1306: Cyclic, Inscribed Quadrilateral, Circle, Tangent, Diagonals, Collinear Points

Subscribe to:
Post Comments (Atom)

Problem 1306

ReplyDeleteIn complete quadrilateral FBECAD the point K is the point Miquel.Then the ΑFCK, ABKD, ABEK and KECD are cyclic.So if x=<FAC=<FKC=<BKE=<BAC=<BDC=<EDC=<EKC.

Therefore the points F,E and K are collinear.But <BKC=<BKF+<FKC=x+x=2x and <BOC=2.<BAC=2x.So <BKC=<BOC or the BKOC is cyclic.But <OBG=90=<OCG or

the BOCG is cyclic with <BOG=<GOC=x.Then the BKOCG is cyclic, so <BKG=<BOG=x=<BKF. Therefore the points F,G,E and K are collinear.

APOSTOLIS MANOLOUDIS KORYDALLOS PIRAEUS GREECE

Assume for a moment that F,G,E are not collinear and so let EG extended meet BF at F'. Let circle BGC meet GE extended at P.

ReplyDeleteSo < GBC = <GCB = < GPB = < BAC = < GPC = < BDC = u say.

So ABEP and CDPE are concyclic.

Further let < F'BG = < BCA < BDA = v say. Then since < BGP = < BCP = v+x, (where < ECP = < EDP = x) and so < BF'G = x.

Hence F'BPD is concyclic and so < BDF' = u.

But < BDF = u, hence F' and F coincide proving that F,G,E are collinear.

Sumith Peiris

Moratuwa

Sri Lanka

Beautiful solution. Very clever. But where it has v+x it must be read v-x.

DeleteMore elegant proof

ReplyDeleteUsing the same notation as before without using F', < ABP = GEC = < PDC hence BFDP is concyclic.

So < FDB = < FPB = < GPB so FGE are collinear

Sumith Peiris

Moratuwa

Sri Lanka

Let m(GBC) = m(GCB) = x

ReplyDelete=> m(BGC) = 180-2x -----(1)

From Alternate segment theorem, we have m(BDC) = m(EDC) = m(BAC) = m(BAE) = x

Join FE and extend it to find a point P such that FCPA and FBPD are concyclic

=> m(FPB) = m(FDB) = x

Similary m(FPC) = m(FAC) = x

Therefore m(BPC) = 2x --------(2)

From (1) and (2) GBPC are collinear and m(GPB) = m(FPB)

Hence F,G,E are collinear

need to prove when FCPA is concyclic, FBPD is concyclic also.

Deleteexcellent solution

ReplyDeletePeter Tran