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.

## Friday, September 30, 2016

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

Labels:
bisector,
chord,
circle,
circumcircle,
concyclic,
cyclic quadrilateral,
exterior angle,
parallel,
triangle

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Let the angles of the triangle ABC be A,B and C respectively.

ReplyDeleteLet 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

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

ReplyDeleteConnect 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

And ∠HFG=∠XBG=w=∠CDF => HF//AD