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 .
Hello thereddevilsI've re-drawn the diagram (attached), so that (which I think is what you meant) looks more like the bisector of .
Then in 's(tangents from a point to a circle)'s are congruent (SAS)
( is angle bisector, given)
is common
Now - and I'll leave you to fill in the details - prove that 's are similar. Hence , and hence .
But (given). Therefore ... ?
Grandad
Hello thereddevilsI have CorelDraw version 12, and Corel Photo-paint 12 on my maching, but they're much too heavyweight for this. So I just use the drawing tools in Word, which are very quick and easy to use. Any text I put in a text box, with no border or background colour. Then, with the image on screen, I use a screen-capture program (MWSnap - freeware, donations appreciated; works fine in Vista) to capture the image to clipboard; paste into Paint; save and then crop in Windows Photo Gallery. Sounds complicated, but it's actually very quick. Graphs I tend to draw in Excel, copy, paste, crop and upload.
Grandad
originally posted by thereddevils
Hello reddevil and grandpa'
The diagrames so far do not appear correct. The initial data are not adequate
Try this
O is the center of the larger circle.The smaller circle passes thru 0
When SP=PV the diagram is unique (only one)Other diagrams SP is not equal PV but always PV=VW andPQ=QR
bjh