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

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

Click the figure below to view more details of problem 1264.

## Saturday, September 24, 2016

### Geometry Problem 1264 Elements: Triangle, Exterior Angle Bisector, Circumcircle, Circle, Perpendicular, 90 Degrees, Concurrent Lines

Labels:
bisector,
circle,
circumcircle,
concurrent,
exterior angle,
perpendicular,
triangle

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To Antonio

ReplyDeleteReferring to problem 1264, The line DF cut arc AC at 2 points G1 and G2.

I understand that the concurrency of BH, EG and AC will be true for both points G per the problem statement.

Do I understand it correctly ? Please clarify.

Peter Tran

To Peter,

DeleteAffirmative answer. The concurrency of BH, EG and AC will be true for both points G. Thanks.

Let AC, EG cut at X

ReplyDelete< DBF = < EBA = < EGA = < ECA = @ say ....(1)

Let < ECB = < EAB = < EGB = € ...(2)

Now < ADB = < € since < BCA = @ + €

So GXBD is concyclic

Hence < XBG = < XDG = 90-@-€

So < BHG = 90

But < XHG = 90 so BXH are collinear points proving that AC, GE and and BH are concurrent

Sumith Peiris

Moratuwa

Sri Lanka

Problem 1264

ReplyDeleteSuppose EG intersects AC at point P, it suffices to show that the BP is perpendicular to AG.

Suppose BK is the internal bisector of <ABC (K=means the arc AC),then KB perpendicular in ED.So arcAE=arcEC .Is <FBD=(<CAB+<ACB)/2=90-<ABC, <GDB=90-<FBD=<ABC/2 and

<CDB=<BCA-<CBD=(<ACB-<BAC)/2. But <PGB=<EGB=<ECB=<ACB-<ACE=<ACB-(<CEB+<CDE)=

=(<ACB-<BAC)/2=<PDB. So the GPBD is cyclic.If BP intersect AG at H’ then <H’PG+<PGA=

=<GDB+<EGA=<ABC/2+<ECA=<KBC+<EKC=90.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

https://goo.gl/photos/NQrpJBvDM9aABBvB8

ReplyDeleteSumith and Apostolis already showed concurrency of BH, EG and AC. I will show that point G1 will have the same property as point G.

Let EG1 meet AC at P and BP meet AG1 at H1

We will show that BP⊥AG1 ( see sketch)

Let u= ∠ (EBA)= ∠ (ECA)= ∠ (EG1A)= ∠ (DBC)

And v=∠ (ECB)= ∠ (EG1B)

In triangle BCD since ∠ (ACB)= u+ v and ∠ (CBD)= u=> ∠ (BDC)= v

In triangle CFD we have ∠ (FDC)= 90-u-v

Since ∠ (PG1B)= ∠ (BDP)=v => BDG1P is cyclic and ∠ (PBG1)= ∠ (PDG1)= 90-u-v

In triangle BH1G1 ∠ (PBG1)+ ∠ (AG1B)= 90-u-v+u+v= 90

So BP⊥AG1