In the parallelogram ABCD, a (1,1) AB vector (6,0) m is the midpoint of line AB, and the intersection of line cm and BD is p I know that the trajectory of P is a circle, but I don't know how to prove it

In the parallelogram ABCD, a (1,1) AB vector (6,0) m is the midpoint of line AB, and the intersection of line cm and BD is p I know that the trajectory of P is a circle, but I don't know how to prove it

Because | ad | = | ab | = 6, set the coordinates of point D as (x0, Y0), and get the coordinates of point C as (x0 + 6, Y0), and
X0=1+6sinθ
Y0=1+6cosθ
Then the linear equation of ad is
（x-7）/(y-1)=(x-x0)/(y-y0) （1）
The coordinate of point m is (4,1), so the equation of PM is
（x-4）/(y-1)=[x-(x0+6)]/(y-y0) （2）
Two straight lines intersect at point P, that is to find the solution of equations (1) and (2)
obtain
x-5=2sinθ
y-1=2cosθ
(x, y) is the coordinate of P, so the trajectory of P is
(x-5)2+(y-1)2=4

In the parallelogram ABCD, a = (1,1), vector AB = (6,0), ad (3,5) point m is the midpoint of line AB, line cm and BD intersect at point P

Connect AC to BD to E,
In the parallelogram ABCD, AE = EC,
Point m is the midpoint of line ab,
The intersection point P of CM and BD is the center of gravity of △ ABC,
∴CP=(2/3)CM.
OA=(1,1),AB=(6,0),AD=(3,5)
The vector AC = AB + ad = (9,5), am = (1 / 2) AB = (3,0),
The vector OC = OA + AC = (10,6), OM = OA + am = (4,1),
The vector cm = om-oc = (- 6, - 5), CP = (2 / 3) cm = (- 4, - 10 / 3),
The vector OP = OC + CP = (10,6) + (- 4, - 10 / 3) = (6,8 / 3),
∴P(6,8/3).