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 5 of 5: Average the normalized coordinates of S and T

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In plain words

Once two points are written in normalized barycentric coordinates (whose entries sum to 1), their midpoint is simply the entry-by-entry average of the two coordinate triples — and here that average is M on the nose.

\dfrac{S+T}{2}=\left(0,\dfrac{a-b+c}{2a},\dfrac{a+b-c}{2a}\right)=\left(0,\dfrac{s-b}{a},\dfrac{s-c}{a}\right)=M

Detailed analysis

In normalized barycentric coordinates, the midpoint of two points is their coordinate-wise average. Averaging $S=(0,\tfrac{a+c}{a},-\tfrac{c}{a})$ and $T=(0,-\tfrac{b}{a},\tfrac{a+b}{a})$ gives $\dfrac{S+T}{2}=\left(0,\dfrac{a-b+c}{2a},\dfrac{a+b-c}{2a}\right)=\left(0,\dfrac{s-b}{a},\dfrac{s-c}{a}\right)=M$ (using $2(s-b)=a-b+c$ and $2(s-c)=a+b-c$ ). Hence $M$ is the midpoint of $ST$ (in fact $MS=MT=s$ ).