2007
Problem - 2337
Let $ ABC$ be acute triangle. The circle with diameter $ AB$ intersects $ CA,\, CB$ at $ M,\, N,$ respectively. Draw $ CT\perp AB$ and intersects above circle at $ T$, where $ C$ and $ T$ lie on the same side of $ AB$. $ S$ is a point on $ AN$ such that $ BT = BS$. Prove that $ BS\perp SC$.
Let $H$ be the intersection point of $CT$ and $AB$. Then, because $AN\perp BC$ and $CH\perp AB$, we find $AHNC$ are concyclic. Hence, by the power of point theorem, we have $BH\cdot BA=BN\cdot BC$.
Meanwhile, in $Rt\triangle{ABT}$, we have $BT^2=BH\cdot BA$. Also considering $BS=BT$, we have $$BS^2=BT^2=BH\cdot BA=BN\cdot BC$$
Now, in $\triangle{BSC}$, because $SN\perp BC$, in order for $BS^2=BN\cdot BC$ hold, $\angle{BSC}$ has to be right which means $BS\perp SC$.