Problem 1

Let

ABC

be a triangle and

J

the center of the

A

-excircle. This excircle is tangent to the side

BC

M

, and to the lines

AB

and

AC

K

and

L

, respectively. The lines

LM

and

BJ

meet at

F

, and the lines

KM

and

CJ

meet at

G

. Let

S

be the point of intersection of the lines

AF

and

BC

, and let

T

be the point of intersection of the lines

AG

and

BC

. Prove that

M

is the midpoint of

ST

Step 3 of 5: Intersect the lines to get F and G

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In plain words

In homogeneous barycentrics, both the line through two points and the intersection of two lines are just cross products of triples, and after factoring out a common nonzero factor both F and G reduce to remarkably clean coordinates.

F=(a:a+c:-c),\quad G=(a:-b:a+b)

Detailed analysis

Forming the line equations $LM$ and $BJ$ via cross products of homogeneous triples and intersecting them (another cross product), then dividing out the common nonzero factor $(s-b)(s-c)$ , gives $F=LM\cap BJ=(a:a+c:-c)$ . The symmetric calculation for $G=KM\cap CJ$ (interchanging $B,C$ and $b,c$ ) gives $G=(a:-b:a+b)$ ; that is, $F=(a:a+c:-c),\quad G=(a:-b:a+b)$ .