Hello,
I just want some input on this proof I did just to get some clarity. Anyway,
I started by extending PT to a point on the circle F, making PF a chord.
Then I extended another line from C that was tangent at F. So CP = CF since they are both tangents from same point.
Since, Triangle FTC and Triangle CTP share 90 degrees and 2 common sides they are congruent (RHS), so PT = TF
CP2 = CB * CA (circle theorem)
and
CT2 = CP2 - TP2
= (CB * CA) - TP2
then,
TP * TF = TA * TB (circle theorem) (Recall PT = TF)
TP * TP = TA * TB
TP2 = TA * TB
Hence,
CT2 = (CB * CA) - (TA * TB)
Q.E.D
Im just not sure about the first step of extending PT to F and then extending the tangent. Is this something you can do?
Thanks
I just want some input on this proof I did just to get some clarity. Anyway,
I started by extending PT to a point on the circle F, making PF a chord.
Then I extended another line from C that was tangent at F. So CP = CF since they are both tangents from same point.
Since, Triangle FTC and Triangle CTP share 90 degrees and 2 common sides they are congruent (RHS), so PT = TF
CP2 = CB * CA (circle theorem)
and
CT2 = CP2 - TP2
= (CB * CA) - TP2
then,
TP * TF = TA * TB (circle theorem) (Recall PT = TF)
TP * TP = TA * TB
TP2 = TA * TB
Hence,
CT2 = (CB * CA) - (TA * TB)
Q.E.D
Im just not sure about the first step of extending PT to F and then extending the tangent. Is this something you can do?
Thanks