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Can any of you offer some advice on how to solve this problem?

Show that if point $P$ is external to a circle and tangent $PT$ and secant $PAB$ are drawn where $T, A$ and $B$ lie on the circle, then $\angle TPA \circeq \frac 12 (BT-AT)$ where $BT$ and $AT$ represent the lengths of the arcs on the circle.

I know that this is a direct consequence of the theorem that states the angle between a chord and the tangent at one of its endpoints is equal in degrees to half the subtended arc. I'll probably need to use this somewhere in the proof, but I'm not sure where.

I appreciate any help.

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Your equation looks as if there is an angle on the left and a length on the right?? – Hagen von Eitzen Feb 7 '13 at 17:17
Yes, sorry. BT and AT are meant to represent the lengths of the arcs on the circle. – brdcastguy Feb 7 '13 at 17:30
It is still nonsense. Let's assume the statement holds for some configuration $P,T,A,B$. Then zoom out the whole picture from $P$: it preserves tangent and section points and the angle $\angle TPA$, but increases the arc lengths $AT$ and $BT$. – Berci Feb 7 '13 at 18:05

enter image description here

$\angle PTA=\angle TBA, \angle TAB=\angle TPA+\angle PTA, \angle TPA=\angle TAB-\angle TBA. $

$\angle TDB=2\angle TAB, \angle TDA=2\angle TBA. \angle TPA=\dfrac{\angle TDB-\angle TDA}{2}$

let $DT=DB=DA=r$

$\angle TDB=\dfrac{\stackrel{\frown}{TB}}{r},\angle TDA=\dfrac{\stackrel{\frown} {TA}}{r}, \angle TPA=\dfrac{1}{2r}(\stackrel{\frown}{TB}- \stackrel{\frown}{TA}).$

so you miss the r .

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