Hi I am currently stuck on this question If anybody could tell me how to approach this or offer any help then please post. Here is the question: Thanks for any help! If you need any more clarificaiton then please just ask SORRY! the point P is the intersection of the chord AC with chord BD
Last edited by Geometor; Dec 22nd 2007 at 08:47 AM.
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You got B and C the wrong way round. Anyway, the problem is the same as saying that 3×∠BOD = ∠COA. ∆POD is isosceles, so ∠POD = ∠PDO. ∴ ∠OPC = ∠POD + ∠PDO = 2×∠POD ∆COD is isosceles, so ∠OCD = ∠ODC = ∠POD. ∴ ∠COA = ∠OPC + ∠OCD = 2×∠POD + ∠POD = 3×∠POD.
Originally Posted by JaneBennet You got B and C the wrong way round. Well, if I had noticed that I wouldn't have spent an idle 20 minutes trying to do it! Good spot. -Dan
Thanks Jane! but why does the labelling of a point matter so much? :S its just a label
Originally Posted by Geometor Thanks Jane! but why does the labelling of a point matter so much? :S its just a label Because the condition 3 BD = AC will be different for your figure
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