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 1 of 5: Tangent lengths turn BJ and CJ into perpendicular bisectors

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In plain words

Tangent segments from the same outside point to a circle are always equal, so B and J are both equally far from K and M, and equally far points from both endpoints of a segment always sit on its perpendicular bisector.

BK=BM=s-c,\quad CL=CM=s-b,\quad AK=AL=s

Detailed analysis

Let $s=\tfrac12(a+b+c)$ with $a=BC,b=CA,c=AB$ . Tangent segments from $B$ to the $A$ -excircle are equal, so $BK=BM$ , and the standard tangent-length formulas give both equal to $s-c$ ; likewise $CL=CM=s-b$ , and $AK=AL=s$ . Since also $JK=JM=JL$ (all radii of the excircle), both $B,J$ are equidistant from $K,M$ , so line $BJ$ is the perpendicular bisector of $KM$ ; similarly line $CJ$ is the perpendicular bisector of $LM$ .