
Geometry Geometry Math Forum 
 LinkBack  Thread Tools  Display Modes 
November 17th, 2015, 11:15 AM  #1 
Newbie Joined: Jul 2011 Posts: 4 Thanks: 0  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 . 
November 17th, 2015, 01:29 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 19,884 Thanks: 1835 
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, 07:49 PM  #3  
Newbie Joined: Jul 2011 Posts: 4 Thanks: 0  Quote:
Last edited by skipjack; November 17th, 2015 at 08:51 PM.  

Tags 
bisecting, lines, points, proof, simple 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Finding tangent lines on functions with given points  cacophonyjm  Calculus  10  March 12th, 2012 02:14 AM 
Coordinate Proof... Perpendicular Lines  julian21  Algebra  2  November 15th, 2010 04:22 PM 
Perpendicular lines proof  Dlarah  Algebra  3  June 17th, 2009 02:55 PM 
Simple proof  temp_tsu  Applied Math  1  September 21st, 2008 04:59 PM 
proof involving points of inflections....  zomgcopters  Calculus  5  October 22nd, 2007 07:43 PM 