P is a point inside a square whose side has a length of 16. P is equidistant from two adjacent vertices and th

Question

its on e-zgeometry.com

Answer 1

Though your question is incomplete,I think you want to know the position of point P so that it is equidistant not only from the opposite vertices but also from the side opposite to these vertices
let ABCD be a square with each side 16 units.M and N are the mid-points of AB and CD respectively.As the point P is equidistant from A and B ,it must be on MN
Let AP=PB=PN=x(say)
Therefore,PM=MN-NP=16-x
Now, In right-angled triangle APM
PM^2+AM^2=AP^2
=>(16-x)^2+8^2=x^2 [as AM=AB/2,M being the mid-pt. of AB]
=>256-32x+x^2+64=x^2
=>-32x=-64-256= -320
=>x=10
Therefore AP=PB=PN=10 units
Hence P will be 10 units from N or 6 units from M on the line segment MN

Answer 2

Too bad you didn't give us the whole question, or someone may have been able to answer it. As it is now, it is unanswerable.