## Wednesday, September 28, 2016

### Geometry Problem 1267 Triangle, Incircle, Excircle, Circle, Tangency Points, Perpendicular, 90 Degrees, Parallelogram

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 1267. 1. https://goo.gl/photos/1anav7EpeN3Uvv8e9

Let ED meet AC at P and E1D1 meet AC at P1
Note that triangles BED and BE1D1 are isosceles and ED//E1D1
CF=AF1= p- AB where p= semi perimeter of triangle ABC
1. Per the result of problem 1265 and 1266 we have
FC/FA= PC/PA= k
We can calculate FP in term of k and AC
FP= ( (k/1+k) + k/(1-k)). AC
Similarly F1A/F1C=P1A/P1C= k and F1P1= ( (k/1+k) + k/(1-k)). AC
So FP=F1P1
Triangle PGF congruent to P1G1F1 ( case ASA) => FG=F1G1
We also have FG//F1G1 => FGF1G1 is a parallelogram

2. we have triangle AFG congruent to CF1G1 ( case SAS) and triangle CGF1 congruent to AG1F1 ( case SAS)
So AG= CG1 and CG= AG1
And AGCG1 is a parallelogram

2. Problem 1267
Is AF_1=CF(O=incircle, O_1=excircle corresponding to AC) or AE_1=AF_1=CF=CD. But triangleAE_1G_1 is similar with triangle G_1D_1C then E_1G_1/G_1D_1=AE_1/CD_1=DC/CD_1 so than vice versa Thales' theorem we have
E_1D//G_1C. Also triangle AEG is similar with triangle GDC so EG/GD=AE/DC=AE/AE_1
then AG//DE_1.Terefore AG//CG_1.Similar CG//AG_1 so AG_1CG is parallelogram.
If AC Intersects GG_1 in M then G_1M=MG and F_1M=MF (AF_1=FC) so F_1G_1FG is
parallelogram.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

3. With the usual notation, 2s = a+b+c,

AE1= AF1 = CF =CD = s-c and
AE = AF = CF1 =CD1 = s-a

From Problems 1265 and 1266 we have

<E1AG = < G1CD1 and < EAG = < DCG, so < GAG1 = GCG1
Further AG1/CG1 = AF1/CF1 = (s-c)/(s-a) = CF/AF

Hence Tr.s GAG1 and GCG1 are similar. However the side GG1 is common to
these 2 triangles. Hence they must also be congruent and therefore AGCG1 is a
parallelogram.

Hence Tr.s AF1G1 ≡ CGF (SAS),
so AG1 = GC
Similarly Tr.s AF1G ≡ CFG1 (SAS),
so AG = CG1

So in quadrilateral FGF1G1, the opposite sides are equal, hence the same
is a parallelogram.

Sumith Peiris
Moratuwa
Sri Lanka