## Monday, September 12, 2016

### Geometry Problem 1260: Triangle, Internal Angle Bisector, Circumcircle, Circle, Perpendicular, 90 Degrees, Concurrent Lines

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 1260.

1. https://goo.gl/photos/Zhmtis9G6kkChPJt7

Connect AF, FC, CG and BG
Let FG meet AC at K
Observe that F is the midpoint of arc AC and GF is the bisector of angle AGC.
We have v= ∠ (ABF)= ∠ (FBC)= ∠ (AGF)= ∠ (FGC)= ∠ (FAC)= ∠ (FCA)
And u= ∠ (HBC)= ∠ (HGE) ( see sketch)
Note that qua. BLMG is cyclic so ∠ (BLG)= ∠ (HGE)= z
We have ∠ (BKD)= ½( Arc AF+Arc BC)
∠ ((BGF)=1/2(Arc AF+Arc AB)= ½(Arc FC+ Arc AB)
Adding above expressions and see that ∠ (BKD) supplement to ∠ (BGF)
So qua. BGKD is cyclic.
In circle BGKD, angle DGF= u- v and arc DK= 2(v-u)
Angle DBH= u-v so BH will meet circle BGKD at K’ where arc DK’= 2(v-u)
So K coincide to K’ and BH, AC and FG are concurrent

2. Let FG, BH intersect at X and let AC, BH intersect at Y

BHEG and BDXG are both concyclic hence < HED = < GDX = 90-C

But < GDC = 90-C

Therefore X and Y must coincide

Hence AC, BH and FG are concurrent

Sumith Peiris
Moratuwa
Sri Lanka

3. Reverse Method

If we assume that H is just the intersection of BX and AG where X is the point of intersection of AC and FG then

< GXC = < GAX + < AGX = < DBG.

So ADXG is concyclic and so < GBC = < GDC = 90-C and since < BGA = C, < BHG = 90.

4. Problem 1260
Let GF intersects the AC in P. It suffices to show that the BP is perpendicular to AG.
Is arcAF=arcFC, <BDA=(arcAB+arcFC)/2=(arcAB+arcAF)/2=(arcBAF)/2=<BGF.So
BDPG is cyclic.Then <BPD=<BGD or <ACB+<PBC=<BGA+<AGD or <PBC=<AGD or
<HBE=<HE.So BHEG is cyclic.Therefore <BHG=<BEG=90.
APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

5. Let us denote the intersection of AG & BF as I, intersection of BC and AF as J, intersection of BC and GF as K and the point of concurrency of BH,FG,AC as P.

Let us denote the angles of triangle ABC as A,B,C respectively.

Since Angle BHG = Angle BEG = 90
=> B,H,E,G are concyclic
=> Angle HBE = Angle HGE ----(1)

Join GC which gives the Angle AGC = ABC = B
=> Angle EGF = Angle HBF ------ (2)

From (1) and (2) we conclude that Angle HGF = Angle JGK = B/2 => Angle KGC = FGC = B/2
and the triangles IBJ and KGJ are similar
Denote Angles of triangle IBJ as I,B/2,J respectively
=> Angle of triangle KGJ are I,B/2,J from similarity

Consider the triangle CKH where Angle CHK = 180-C-I -------- (3)
Consider the triangle ABK where Angle AHB = 180-A-(B/2+90-I) = 90-A-B/2+I ----------(4)

Now Join BG; Angle AGB = Angle ACB = C
Since Triangle BHG is right triangle => Angle HBG = 90-C

Now consider the triangle PBG
Angle BPG = 180-(Angle PGB + Angle PBG)
=> BPG = 180 - (B/2+C+90-C)
=> BPG = 90 - B/2 ------------ (5)

=> CHK+AHB+BHG = 180-C-I+90-A-B/2+I+90-B/2 = 180+180-(A+B+C) = 180

Hence BH,FG & AC are concurrent

6. Slight correction to my previous solution. I have denoted H instead of P: Consider below solution:

Let us denote the intersection of AG & BF as I, intersection of BC and AF as J, intersection of BC and GF as K and the point of concurrency of BH,FG,AC as P.

Let us denote the angles of triangle ABC as A,B,C respectively.

Since Angle BHG = Angle BEG = 90
=> B,H,E,G are concyclic
=> Angle HBE = Angle HGE ----(1)

Join GC which gives the Angle AGC = ABC = B
=> Angle EGF = Angle HBF ------ (2)

From (1) and (2) we conclude that Angle HGF = Angle JGK = B/2 => Angle KGC = FGC = B/2
and the triangles IBJ and KGJ are similar
Denote Angles of triangle IBJ as I,B/2,J respectively
=> Angle of triangle KGJ are I,B/2,J from similarity

Consider the triangle CKP where Angle CPK = 180-C-I -------- (3)
Consider the triangle ABP where Angle APB = 180-A-(B/2+90-I) = 90-A-B/2+I ----------(4)

Now Join BG; Angle AGB = Angle ACB = C
Since Triangle BHG is right triangle => Angle HBG = 90-C

Now consider the triangle PBG
Angle BPG = 180-(Angle PGB + Angle PBG)
=> BPG = 180 - (B/2+C+90-C)
=> BPG = 90 - B/2 ------------ (5)