## Monday, January 2, 2012

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

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

Click the figure below to see the problem 711 details.

1. Let E' be the midpoint of AD. Join EE'
BA ∥ 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

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

Draw 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

3. 1 and 2 are easily proved considering congruent triangles.

Let 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

4. Problem 711
The 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