ABCD is a square with AB=12. Point P is in interior and the distances to A,B and to the side CD are equal. Find the distance.

Let P be  the point which is equidistant from A, B and
CD.

Then the point P being equidistant from A and B, the
perperpendicular bisector of AB shoud pass through P  and bisect AB at E and CD
at F respectively.

Now  the triangel AFE is a right angled triangle.
Therefore AF^2 = AE^2+EF^2.

AF^2= (1/2)AB^2 + EF^2,  as E is mid
point of AB.

AF^2 = 6^2 + 12 ^2 , as EF = BC = CD = 12 , sincw ABCD
is a square.

Therefore AF^2 =  36+144 =
180.

Therefore AF = sqrt(180) =
6sqrt5.

Similarly BF = 6sqrt5.

AB = 12
given

The area of triangle ABF = (1/2)AB*FF = (1/2)12*12 =
72.

Therefore AP = BP = FP = R the circumradius  R of the triangle
ABF , where F is the mid point of CD.

We know that in a triangle
circumradius R =   abc/4*area of trangle .

Therefore the circum
radius of ABF is R = AB*AF*BF/4*area of triangle = 12*(6sqrt5)(6sqrt5)/(4*72) = 12*36*5/(4*72) =
7.5cm.

Therefore the  distance AP= BP = FP =
7.5cm.