My Math Forum Simple(?) proof on points on bisecting lines.

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November 17th, 2015, 10:15 AM   #1
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Simple(?) proof on points on bisecting lines.

Hey all,

I'm an adult going back through geometry (the McDougal text). Only in chapter 4, but it's going smoothly. However, there's one particular problem that is tripping me up - it seems like I'm supposed to assume something that isn't necessarily true. Either that or I'm missing the obvious, but I've asked some other people and they don't see anything either.

Any help would be very appreciated .
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 November 17th, 2015, 12:29 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,373 Thanks: 2010 Choose F on BD extended, G on BE extended and H on DE so that $\angle$DFP = $\angle$DHP = $\angle$EGP = 90°. As triangles PFD, PHD are congruent, PF = PH. As triangles PGD, PHD are congruent, PG = PH. Hence PF = PG, and so triangles PFB, PGB are congruent, which implies that $\angle$PBF = $\angle$PBG, i.e. that BP bisects $\angle$ABC.
November 17th, 2015, 06:49 PM   #3
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Quote:
 Originally Posted by skipjack Choose F on BD extended, G on BE extended and H on DE so that $\angle$DFP = $\angle$DHP = $\angle$EGP = 90°. As triangles PFD, PHD are congruent, PF = PH. As triangles PGD, PHD are congruent, PG = PH. Hence PF = PG, and so triangles PFB, PGB are congruent, which implies that $\angle$PBF = $\angle$PBG, i.e. that BP bisects $\angle$ABC.

Last edited by skipjack; November 17th, 2015 at 07:51 PM.

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