AB and CD are two chords of a circle with centre O, and the two chords are equal in length. AB is produced beyond B and CD is produced beyond D, and the two produced lines meet at a point P outside the circle. Which statement must be true?

Write the common chord length as $L$, so PA = $PB+L$ and PC = $PD+L$. The power of the point $P$ is the same along both lines: PA × $PB$ = PC × $PD$, that is $PB(PB$ + L) = $PD(PD$ + L). Rearranging, $P{B}^2-P{D}^2+L(PB-PD)=0$, so $(PB-PD)(PB+PD+L)=0$. The second bracket is positive, so $PB=PD$ (and hence PA = PC as well). PA = $PB$ would force $B$ to coincide with A, and OP bisects ∠APC rather than meeting AB at a right angle — that would need OP to run through the midpoint of the chord, which it does not.