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November 17th, 2015, 11:15 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, 01:29 PM   #2
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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.
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November 17th, 2015, 07:49 PM   #3
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Quote:
Originally Posted by skipjack View Post
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.
Ah, perfect! I had made the altitude from P to BD/BE (your F and G), but I hadn't thought about making an additional one to DE - that was the missing piece. Thanks!

Last edited by skipjack; November 17th, 2015 at 08:51 PM.
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