I've never heard it called that but, as you will see from this link, I did know this:
http://www.mathisfunforum.com/viewtopic … 92#p368592
Bob
]]>PDA
PAB
So PB/PA involves taking the first and third letters from the bottom three and the first and third from the top three. I've underlined the ones I mean.
So I can set that equal to another ratio taken in the same way. As I want DA and AB in the final answer, it makes sense to take them; so:
So that's half way there.
Now your turn.
Repeat for the other pair of similar triangles, again looking to make PB/PA. Except there's no PA this time. No worries; because the two tangents are equal, so you can use PB/PC.
Bob
ps. I've never heard of "power of a point". Would you care to explain when you get time?
]]>I'll be on this forum fairly actively if I need any homework help.
]]>Welcome to the forum.
Homework due again?
Backgound reading here:
http://www.mathisfunforum.com/viewtopic.php?id=22506
Hint: Use the angle between a chord and a tangent is equal to the angle made by the chord to show angle BDC = angle BCP and hence that triangles PCB and PDC are similar.
Do a similar thing for triangles PDA and PAB.
Also note that PA = PC.
Writing ratios for PB/PA should enable you to complete this.
Bob
]]>I've already tried Power of a Point and similar triangles, but I can't seem to get anything to work. Can someone help me directly? By the way, don't post another link from this website, as I've probably checked that one already. Many thanks!
P.S. I need the answer before Saturday night.
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