## Friday, September 30, 2016

### Geometry Problem 1268:Triangle, Exterior Angle Bisector, Circumcircle, Circle, Chord, Parallel, Concyclic Points, Cyclic Quadrilateral

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click on the figure below to view more details of problem 1268.

1. Let the angles of the triangle ABC be A,B and C respectively.
Let Angle BFD = @ => BFG = 180-@ -----(1)

Since ED is the ext.Angle bisector of B ,Angle CBD = 90-B/2
=> Angle BDF = 90+B/2-@

Angle ABE = 90-B/2 = AGE = ACE => ECB = C+B/2-90 = XDB = EGB = XGB
Hence BPGD is concyclic

Since BXG and BDG are supplimentary, Angle EXB = 90+B/2-@ = HXG
Hence in triangle HXG, Angle XHG = BHG = @ ---------(2)

From (1) & (2) BHGF are concyclic --------(3)

Since Angle ACB = C = AGB = HGB = C => HFB = C (From (3) ) Hence HF // AC // AD

2. https://goo.gl/photos/8vPumpKGSCYHa6DE9

Connect EC and BG
Let u= ∠CBD=∠EBA=∠ACE=∠AGE
And v= ∠ECB= ∠EGB ( see sketch)
In triangle BDC , ∠BDC=∠BCA-∠CBD= v
Since ∠BDX=∠XGB= v => quadrilateral BDGX is cyclic
And ∠ XDG=∠XBG= w
In triangle BFD, ∠BFD= 180-u-v-w
In triangle HGB , ∠BHG= 180-u-v-w
So ∠BHG=∠ BFD and quadrilateral HGFB is cyclic