9.3 Equal Chords and Distance from Centre

Construction: Draw $OM \perp AB$ and $ON \perp CD$. Join $OP$.

Since $AB=CD$, they are equidistant from centre. So $OM=ON$ (Thm 9.6).

In $\Delta OMP$ and $\Delta ONP$: $OM=ON$ $\angle M = \angle N = 90^\circ$ $OP=OP$ (Common Hypotenuse).

By RHS, $\Delta OMP \cong \Delta ONP$. Thus $MP = NP$.

We know $AM = AB/2$ and $CN = CD/2$. Since $AB=CD$, $AM=CN$.

Add equations: $AM + MP = CN + NP \implies AP = CP$.

Subtract from totals: $AB - AP = CD - CP \implies BP = DP$.