Q.In Fig. 1.15, the point D is chosen such that we have to extend OC for the perpendicular DN to fall on it. What will happen to the final expression of I₀, if point D is so chosen that the perpendicular DN falls directly on OC?

Nothing changes — the theorem's result is independent of where the mass element sits. …

In the proof, $(DO)^2 = (DN)^2 + (NO)^2$ with $NO = NC + h$ when N lies beyond C (OC extended). If D is instead chosen so that the foot N falls BETWEEN O and C, then $NO = h - NC$, and expanding gives $(DO)^2 = (DC)^2 - 2h(NC) + h^2$ — the cross term now enters with a MINUS sign. But NC is precisely a signed coordinate of the mass element measured from the centre of mass along the OC direction: integrated over the whole body, $\int NC\,dm = 0$ by the very definition of the centre of mass, regardless of sign conventions — elements on the two sides of C cancel. So the middle term …

Redo the two-line expansion with a signed coordinate x = NC measured positively from C towards O: $(DO)^2 = (DN)^2 + (h - x)^2$, integrate, and watch $\int x\,dm = 0$ kill the cro …

Believing the theorem needs every element's perpendicular foot beyond C (it doesn't — a real body has elements on BOTH sides, and the proof already relies on their can …