Hello thereddevils Originally Posted by
thereddevils In the figure , ST is the common tangent at P to the two circles . PVW and PQR are two straight lines . OS bisects \angle PSR and PV=PS , the lines QV and RW are also parallel .
Prove that SR=VW .
I've re-drawn the diagram (attached), so that $\displaystyle QS$ (which I think is what you meant) looks more like the bisector of $\displaystyle \angle PSR$.
Then in $\displaystyle \triangle$'s $\displaystyle SQP, SQR$$\displaystyle SP = SR$ (tangents from a point to a circle)
$\displaystyle \angle PSQ = \angle SRQ$ ($\displaystyle SQ$ is angle bisector, given)
$\displaystyle QS$ is common
$\displaystyle \therefore \triangle$'s $\displaystyle SQP, SQR$ are congruent (SAS)
$\displaystyle \therefore PQ=RQ$
Now - and I'll leave you to fill in the details - prove that $\displaystyle \triangle$'s $\displaystyle PVQ, PWR$ are similar. Hence $\displaystyle PV:VW=PQ:QR$, and hence $\displaystyle PV=VW$.
But $\displaystyle PV = PS$ (given). Therefore ... ?
Grandad