Given a square ABCD with side 8.A circle is drawn through vertices A and D and tangent to the side BC.find the radius of the circle.
Suppose M is the point where the circle tangents BC. BM=MC. Let's put point N on AD, AN=ND. MN is the height of the triangle AMD. Let O is the center of the circle, then AO=MO=DO. From the triangle AON, where ON is perpendicular AN, AO^2=AN^2 ON^2. If r is the radius of the circle, then AO = r, AN = AD/2 = 4, ON = 8 - r (MN=AB=8, MO = r). So, (8 - r)^2 4^2 = r^2, 64 - 16r r^2 16 = r^2, 16r = 80, r = 5