April wrote: 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 $ CT$ and $ AB$ intersect at $ K$. Since $ AB$ is diameter of the circle and $ N$ lies on that circle, so $ \angle{BNA}=90$. Also we have that $ \angle{BKC}=90$ so triangles $ BKC$ and $ BNA$ are similar. Hence, $ \angle{BCK}=\angle{BAN}$. We also have that $ \angle{BTA}=90$, because $ T$ lies on the circle. Hence we have that triangles $ BKT$ and $ BTA$ are similar thus $ \frac{BT}{BK}=\frac{BA}{BT}$ or $ BT^{2}=BA.BK$. Since $ BT=BS$ we have $ BA.BK=BS^{2}$ hence $ BS$ is tangent to circle circumscribed around $ KSA$. It follows that $ \angle{BSK}=\angle{KAS}=\angle{BAN}=\angle{BCK}$. From here we get that $ BCSK$ is cyclic and thus $ \angle{BSC}=\angle{BKC}=90$, in other words $ BS\perp SC$.

let$ H$ be feet of $ C$. $ \angle T=90^{\circ}$ hence $ BS^{2}=TB^{2}=BH.BA=BN.BC$ and cause $ \angle N=90^{\circ}$ hence $ \angle S=90^{\circ}$.

Just as $T$ is the intersection of the circle with diameter $AB$ with the height from $C$, we wish to show that $S$ is the intersection of the circle on $BC$ with the height from $A$. Let's rather redefine $S$ and $T$ as the intersection of a height and a Thales circle of a side, and show that $BS=BT$. Now note that if $X$ is the midpoint of $BC$, and $BN=n=c\cos \beta$, $XB=XS=r=\frac a2$, then we have $$BS^2=BN^2+NS^2=BN^2+(SX^2-NX^2)=n^2+r^2-(n-r)^2=2nr=ac\cos \beta,$$ which is symmetric in $a,c$, so $BT^2$ will be the same expression. This verifies $BS=BT$.

Hi! Here's my solution

Can't believe this problem is Chinese. First notice that $\angle BTN=\angle BAN=\angle TCN$ Thus ,we get $BT$ is tangent to $\odot(NTC)$, So by power of a point $BT^2=BN.BC=BS^2$ Thus $\triangle BSC$ and$\triangle BNH$ are similar, so we are done.

Let CT meet AB at K. ∠ATB = ∠ANB = 90 so BT^2 = BK.BA = BN.BC = BS^2 so ∠BSC = ∠BNS = 90. we're Done.