Plane vector arithmetic proposition

This problem is very suitable for vector method.

Let the center of the circle be O, the radius be R, and the normal vectors OA, OB, OC and OD be A, B, C and D, where A? = b？ = c？ = d？ = r？ .

P, A and B are collinear, OP = ta+( 1-t)b is collinear with P, C and D, OP = SC+(1-s) D.

So ta+(1-t) b = sc+(1-s) d1.

The deformation is (t/(t-s)) a+(-s/(t-s)) c = ((1-s)/(t-s)) d-((1-t)/(t-s)) B2.

From the left end, we know that the endpoint of the vector is on the straight line AC, and from the right end, we know that the endpoint of the vector is on the straight line BD, so the vector is OE.

Similarly, of = (t/(t+s-1)) a+((s-1)/(t+s-1)) d = (s/(t+s-1)).

Consider (t-s) OE op = t? Answer? -sta c+( 1-t)( 1-s)b d-( 1-t)？ d？ = (2t- 1)r？ -sta c+( 1-t)( 1-s)b d。

(2) Square t on both sides of the formula? Answer? +s？ c？ -2sta c = ( 1-s)？ d？ +( 1-t)？ b？ -2( 1-t)( 1-s)b d。

so-sta c+( 1-t)( 1-s)b d =(( 1-s)？ d？ +( 1-t)？ b？ -t？ Answer? -s？ c？ )/2 = ( 1-s-t)r？ .

Get OE OP = r instead? Similarly, OP = r? .

If m, e and f are collinear, let OM = uOE+( 1-u)OF, then om op = uoe op+( 1-u) of op = r? .

FrOM the m on the circle, there is om? = r？ So pm om = (om-op) om = om? -r？ = 0.

So the straight line PM is perpendicular to the outer end of the radius OM, that is, PM is the tangent of the circle.