$AB$ is a chord (not a diameter) of a circle with centre $O$. Let $T$ be a point on segment $OB$. The line through $T$ perpendicular to $OB$ meets $AB$ at $C$ and the circle at $D$ and $E$. Denote by $S$ the orthogonal projection of $T$ onto $AB$ . Prove that $AS \cdot BC = TE \cdot TD$.
Problem
Source: Pan African MO 2012 Day 1 Q1
Tags: geometry proposed, geometry
ACCCGS8
14.10.2012 16:06
Let $OB$ meet the circle again at $F$. Then $\angle FAB = 90$ and $\triangle TSB$ is similar to $\triangle FAB$. Also $FACT$ is cyclic. These observations yield: $\frac{TF}{TB} = \frac{AS}{BS}$ (1), $BC \cdot BA = BT \cdot BF$ (2) and $\frac{BF}{TB} = \frac{BA}{BS}$ (3). Multiplying (1) and (2) gives $AS \cdot BC \cdot BA = TF \cdot BF \cdot BS \implies AS \cdot BC = \frac{TF \cdot BF \cdot BS}{BA} = TF \cdot TB \cdot \frac{BF}{TB} \cdot \frac{BS}{BA} = TF \cdot TB = TE \cdot TD$. The last part follows from (3).
somebody
12.12.2012 15:23
We know that $ TD=TE $, then we have to prove that $ AS\cdot BC=TD^2 $ $ AS\cdot BC =AC\cdot BC+CS%Error. "clod" is a bad command. BC =CE\cdot CD+TC^2 =(TD+TC)(TD-TC)+TC^2 =TD^2-TC^2+TC^2 =TD^2 $
sunken rock
13.12.2012 15:15
Similar to first solution: $AFTC$ cyclic implies $BT\cdot BF=BC\cdot AB\ (\ 1\ )$, power of $T$: $BT\cdot FT=TD\cdot TE\ (\ 2\ )$; divide the relations $(1):(2)$ and get $\frac{BF}{FT}=\frac{BC\cdot AB}{TD\cdot TE}\ (\ 3\ )$, but $TS\parallel AF\implies \frac{BF}{FT}=\frac{AB}{AS}\ (\ 4\ )$. From $(3)$ and $(4)$ we get the required relation. Remark: $BF$ can be any other chord, $\overline{DTE}$ an antiparallel of $AF$, and $TS\parallel AF$. Best regards, sunken rock
andrejilievski
20.08.2013 23:32
Similar to first and third solution We have $ TE\cdot TD=BT\cdot TF $ so we need to prove that $ \frac{AS}{TF}=\frac{BT}{BC}=sinBCT $ so it is enough to show that $ \frac{AS}{TF}=sinBCT, \frac{AS}{TF}=\frac{SB}{BT}=sinSTB=sinBCT, QED $
Itticantdomath
14.04.2016 17:38
AS.BC =AC.BC+BC.CS AC.BC is the power of C. SO, AC.BC= EC.CD= (ET+CT)(DT-CT) DT=ET. So, AC.BC= (ET.ET)-(CT.CT) Let us examine the power of C from the circle BST. As T is tangent to that circle, CT.CT=BC.CS SO, AS.BC=(ET.ET)-(CT.CT)+(CT.CT)=ET.ET=ET.DT
EulersTurban
25.03.2020 00:08
So first of all let's add a circle around the $\triangle STB$. By the power of point we know that: $CS.CB=CT^2$ and $DC.CE= AC.CB$ By the following procedure we get the following: $$DC.CE=AC.CB$$$$(DT-CT)(ET+TC)=(AS-CS)BC$$$$TD.TE+TD.TC-CT.ET-CT^2=AS.BC-CS.BC$$$$TD.TE+TC(DT-ET)=AS.BC$$Because $\angle DTO = \angle ETO = 90$, and we have that the points $B,T,O$ are all collinear. We get that $BO$ is the perpendicular bisector of $DE$,thus $DT=ET$ Which gives us $AS.BC=TE.TD$.....