Geometry Problem

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

Click the figure below to see the problem 711 details.

## Monday, January 2, 2012

### Problem 711: Rectangle, Midpoint, Angle, Concyclic Points, Cyclic Quadrilateral

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Let E' be the midpoint of AD. Join EE'

ReplyDeleteBA ∥ EE' ∥ CD

∆s ABE and DCE are congruent.

So ∠CDE = ∠BAE

∠AED = ∠AEE' + ∠DEE' = ∠BAE + ∠CDE = 2∠BAE.

Similarly ∠AFB = 2∠DAF is proved.

For convenience denote ∠AED = 2x, ∠AFB = 2y

So ∠BAG = x, ∠ABG = 90° - y

∴ ∠EGN = ∠BGA = 90° + y - x

∠GNH = ∠GEH + ∠EGN = 2x + 90° + y - x

= 90° + x + y

Next ∠EAF = 90° - ∠BAE - ∠DAF

= 90° - x - y

Hence ∠GNH + ∠EAF + 180°

and A, G, N, H are concyclic

http://img249.imageshack.us/img249/3880/problem711.png

ReplyDeleteDraw EX//BA and FX//BC ( see picture)

1, 2. We have ( AEX)=(BAE)= a and (AFX)=(FAD)= b ---- (properties of // lines)

EX and EY are angle bisectors of (AED) and (AFB)

So (AED)=2(BAE)= 2a and (AFB)=2(DAF)=2b

3. From the picture we have (EAF)= 90-a-b -------(1)

(FNY)=b+(YXF)

(EXY)=a+(YXE)

(FNY)+(EXY)= a+b+(EXF)=90+a+b= ( GNH) ---- (2)

Add (1)+(2) side by side : (EAF)+(GNH)=180

4. and A,G,N,H are concyclic

1 and 2 are easily proved considering congruent triangles.

ReplyDeleteLet FIH be parallel to AD such that H is on AE and I is on DE

< EAF + < AFH = < EHI = HIE = < DNF + NFI. Now FH bisects < AFB, hence < EAF = DNF from which results 3 and 4 follow easily.

Sumith Peiris

Moratuwa

Sri Lanka

Problem 711

ReplyDeleteThe AF intersects BC at P,then <FPB=<FBP and <HEP=<AEB. Is <NHA=<DHP=<DEP+<HPB=

<AEB+<FBE=<GEB+<GBE=<NGE. Therefore A,G,N,H are concyclic.

MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE