Geometry Problem

Click the figure below to see the complete problem 548 about Triangle, Transversal, Complete Quadrilateral, Circumcircles, Circumcenters, Similarity, Concyclic Points.

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Level: High School, SAT Prep, College geometry

## Thursday, December 16, 2010

### Problem 548: Triangle, Transversal, Complete Quadrilateral, Circumcircles, Circumcenters, Similarity, Concyclic Points

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ReplyDelete1 C1, C2, C3 and C4 meet at a point G : See solution of problem 547

2 O4O2 perpendicular to DG and O2O3 perpen, to GE so m(DGE) =m(B)

And m(O4O2O3)=m(B)

Similarly m(O2O4O3)=m(A) so triangle ABC similar to triangle O4O2O3 .

3 We have O1O2 perpen. To BG and O1O4 perpen to AG

So angle( O4O1O2) supplement to angle (BGA)

But angle(BGA)= angle(C) =angle (O4O3O2)

So 4 points O1,O2,O3 ,O4 concyclic.

We have angle (BGA)= angle (DGF) both angle supplement to angle A

angle (BCF)= angle (DBG) both angle supplement to angle( ACG)

So tri. GBD similar to tri. GCF .

Tri. GCF is the image of Tri. GBD in the rotation +dilation transformation with center at G

Since O2, O3 is the center of circumcircles of tri. GBD and GCF , O3 will be the image of O2 in this transformation and angle (O2GO3)= angle (BGC) = supplement to angle (A) or angle (O2O4O3)

points O1, O2, O3, O4 and G are concyclic

Peter Tran

(1)

ReplyDeleteLet C2 intersect C3 again at G

DE subtends <B at G,

EF subtends <C at G.

So DF subtends <(Pi - A)at G & < A at A;

A,F,G,D are concyclic; C4 passes through G

(2)

O2O3 perpendicular GE,

O3O4 perpendicular GF,

O2O4 perpendicular GD.

<O2O3O4 = <EGF = <C,

<O2O4O3 = Pi - <DGF = <A,

<O3O2O4 = <DGE = <B.

So triangles ABC, O3O2O4 are similar

(3)

O1O2 perp BG,

O1O3 perp CG,

So <O2O1O3 = <A = <O2O4O3;

O1,O2,O3,O4 are concyclic