José wrote: Consider a circle with centre $O$, and $3$ points on it, $A,B$ and $C$, such that $\angle {AOB}< \angle {BOC}$. Let $D$ be the midpoint on the arc $AC$ that contains the point $B$. Consider a point $K$ on $BC$ such that $DK \perp BC$. Prove that $AB+BK=KC$. 1) Let $DK$ cut the circle again at $E$, $DO$ cut the circle again at $F$. $\angle FED = 90$ hence $FE \parallel BC$ 2) $EB = FC = FA$ hence $EA \parallel FB$, $AB = EF$ 3) Draw parallel from $B$ to $DK$. Let $G$ be the point where it intersects $EF$, then $GE = BK$. 4) Triangles $BFG$ and $CKE$ have equal angles and $BF = EC$ hence $\triangle BFG = \triangle CKE$ hence $FG = CK$ and from 2) and 3) $AB + BK = CK$ Daniel

also known as Archimedes Broken Chord