Saturday, June 8, 2013

Problem 886: Right Triangle, Incenter, Angle Bisector, Perpendicular, 45 Degrees, Concyclic Points, Isosceles Right Triangle

Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
Post your solution in the comments box below.

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

Click the figure below to see the complete problem 886.

1. ∵ ΔAEB≡ΔAEG and ΔCDB≡ΔCDF
∴ AB=AG and CB=CF
i.e. ΔABG and ΔCBF are isosceles.

That means the figure is essentially equivalent to that in Problem 792.

That is, if we draw BM⊥AC at M,
then H and L are the incenters of ΔABM and ΔCBM, respectively.

⇒ BF bisects ∠ABM and BG bisects ∠CBM
⇒ ∠FBG=45°

∵ ΔABG and ΔCBF are isosceles
∴ AJ⊥BG and CK⊥BF
∴ BJIK concyclic with diameter BI.

Also, it follows from Problem 792,
that FHILG concyclic, with diameter FG.

∵ FG is the diameter of circle FHILG
∴ FI⊥IG

From Problem 790, IB=IF=IG.
Thus, ΔFIG is isosceles right triangle.

2. Draw altitude BP
1. We have FP/FA=DB/DA=BC/AC = cos(C)=cos(∠ABP)=BP/AB ( property of angle bisector)
So BF is angle bisector of ∠ (ABP) .
Similarly BG is angle bisector of ∠ (CBP) .
So ∠ (FBG)=1/2*(∠ABC)= 45 degrees
2. ∠ (IAC)+ ∠ (ICA)= ½(∠BAC+ ∠ BCA)= 45 so ∠ (AIC)=135 => B, K, I, J is concyclic
3. Note that DC is a perpendicular of BF => BF ⊥DC and LB=LF
Triangles BKl, FKL are isosceles => ∠BLK=∠KLF= 45
So ∠FLG= 90 .
Similarly FH is perpendicular to BH.
4. Since I is the intersection of perpendicular bisector of BF and BG so I is circumcenter of triangle FBG
So IF= IG and % FIG= 90 and F, H, I, L, G is concyclic and FIG is right isosceles triangle