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 4 of 5: Project from A onto BC to get S and T

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In plain words

Intersecting a line through A with the opposite side BC in barycentric coordinates simply zeroes out the A-coordinate while keeping the B- and C-coordinates unchanged.

S=(0:a+c:-c),\quad T=(0:-b:a+b)

Detailed analysis

Because $A=(1:0:0)$ and line $BC$ is given by $x=0$ , intersecting $AF$ with $BC$ simply drops the first coordinate of $F$ , giving $S=AF\cap BC=(0:a+c:-c)$ , and similarly $T=AG\cap BC=(0:-b:a+b)$ ; that is, $S=(0:a+c:-c),\quad T=(0:-b:a+b)$ . Since the coordinate sums of both $(0,a+c,-c)$ and $(0,-b,a+b)$ equal $a$ , dividing by $a$ already puts both $S$ and $T$ in normalized barycentric form (coordinate sum $1$ ), while $M=(0:s-b:s-c)$ normalizes to $(0,\tfrac{s-b}{a},\tfrac{s-c}{a})$ (using $(s-b)+(s-c)=a$ ).