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Parallel to angle bisector

Source: Centroamerican 2012, Problem 6

June 20, 2012

geometryincenterparallelogramgeometry unsolved

Problem Statement

Let

ABC

be a triangle with

AB < BC

, and let

E

and

F

be points in

AC

and

AB

such that

BF = BC = CE

, both on the same halfplane as

A

with respect to

BC

.
Let

G

be the intersection of

BE

and

CF

. Let

H

be a point in the parallel through

G

to

AC

such that

HG = AF

(with

H

and

C

in opposite halfplanes with respect to

BG

). Show that

\angle EHG = \frac{\angle BAC}{2}

.

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