As shown in the figure, in the rectangular coordinate system, one side of an acute triangle AOB coincides with the positive half axis of the X axis, and the other side OA intersects the image of the function y = 1 / X at point P. take point P as the center of the circle, take 2PO length as the radius, draw an arc to intersect the image of y = 1 / X at point R, and make parallel lines of the X axis and Y axis through points P and R respectively to obtain the rectangular pqrm, connecting OM

As shown in the figure, in the rectangular coordinate system, one side of an acute triangle AOB coincides with the positive half axis of the X axis, and the other side OA intersects the image of the function y = 1 / X at point P. take point P as the center of the circle, take 2PO length as the radius, draw an arc to intersect the image of y = 1 / X at point R, and make parallel lines of the X axis and Y axis through points P and R respectively to obtain the rectangular pqrm, connecting OM

1)
Pqrm four point coordinates:
p(xp,1/xp);
q(xp,1/xr);
r(xr,1/xr);
m(xr,1/xp);
The slope of OP is 1 / (XP · XR)
The slope of OM is 1 / (XR · XP)
They have the same slope, and they all pass point o
So OP and OM are in the same line
The point q is on the line OM
2）
If the image of y = 1 / X is arced at point R with 2PO length as radius, then
√(xr^2+1/xr^2)=2√(xp^2+1/xp^2)
And | PR | = √ [(XP XR) ^ 2 + (1 / XP + 1 / XR) ^ 2]
=√[(xp^2+1/xp^2)-2xp·xr-2/(xp·xr)+(xr^2+1/xr^2)]
=√[(3/2)(xr^2+1/xr^2)-2(xp·xr+1/(xp·xr))]
Let the center of the rectangle be t; then it is easy to prove that ∠ PTO = 2 ∠ mqr = 2 ∠ mob
As long as it is proved that PT = Po, or PR = or, that is ∠ PTO = ∠ AOM, it is OK