Geometry Problem

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

Click the figure below to see the complete problem 678.

## Monday, October 17, 2011

### Problem 678: Triangle, Simson Line, Circumcircle, Tangent, Parallel, Perpendicular, Collinear Points

Labels:
circumcircle,
collinear,
line,
parallel,
perpendicular,
Simson,
tangent,
triangle

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http://img818.imageshack.us/img818/2386/problem678.png

ReplyDeleteConnect BE ( see picture)

Note that quadrilateral BGED is cyclic

So ( BGD)=(BED)

But (BED)=(BAM) both angles face the same arc

So (BAM)=(BGD) and AM//GDF

Peter Tran

<MAG

ReplyDelete= <MAB (same angle)

= <ACB (angle in the alternate segment)

= <AEB (angles in the same segment)

= <DEB (same angle)

= <DGB (B,G,E,D are concyclic; angles in the

same segment)

= <DGA (same angle)

So AM // GD

considering usual notations, let A,B and C be the angles of the triangle ABC

ReplyDeletehence m(MAB)=C (alternate-segment theorem)

Therefore m(MAC) = A+C ---------(1)

Since ADC is right triangle, m(DAC) = 90-C => m(BAD)=A-(90-C) = A+C-90

Observe that A,G,E and F are concyclic

=> m(BAD)=m(GAE)=m(GFE)=A+C-90

=> m(GFC)=A+C -----------(2)

From (1) and (2) GDF//AM