Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to enlarge the problem 945.

## Thursday, December 19, 2013

### Geometry Problem 945: Triangle, Concurrent Lines, Cevians, Perpendicular, Angle Bisector

Labels:
angle bisector,
cevian,
concurrent,
perpendicular,
triangle

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http://img15.imageshack.us/img15/4941/dfl0.png

ReplyDeleteLet B’, F’, D’ are the projections of B, F, D over AC

Apply Ceva’s theorem on triangle BFD with Cevians BG, FC and DA

(GF/GD).(CD/CB).(AB/AF)= 1 => (GF/GD)=(CB.AF)/(CD.AB) ……. (1)

Use properties of similar triangles CDD’ to CBB’ and triangles AFF’ to ABB’ we have

FF’/DD’= (FF’/BB’)/(DD’/BB’)=(AF/AB)/(CD/CB)………(2)

We also have GF/GD= HF’/HD’… ( Thales theorem on parallel lines)

From (1) and ( 2) we have GF/GD=FF’/DD’=HF’/HD’ => triangles FF’H similar to DD’H … ( case SAS)

So GF/GD=HF/HD => HG is angle bisector of angle ( FHD)

if X=intersection of lines DF and AC then (A,C,E,X)=1

ReplyDeleteangle GHX is 90 so H lies on Apollonius circle with diameter GX so FG/GD=FH/HD by the

angle bisector theorem we have what we wanted