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April 19th, 2018, 11:32 PM  #1  
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2  Proof that two line segments are equal Quote:
From OP=OQ comes out that the triangle OPQ is isosceles triangle. This means that $\displaystyle \measuredangle OPQ = \measuredangle OQP$. Now, $\displaystyle \measuredangle AQP = \measuredangle AQO + \measuredangle OQP = 90$/degree$ + \measuredangle OQP$ and $\displaystyle \measuredangle DPQ = \measuredangle DPO + \measuredangle OPQ = 90$\degree$ + \measuredangle OPQ$. So, $\displaystyle \measuredangle AQP = \measuredangle DPQ$. This means that quadrilateral EDPQ is isosceles trapezoid. This means that $\displaystyle EQ = DP$. That's it. I can't go further. P.S. Anybody have the idea how to write degrees sign in Latex? I copied this from online Latex equation editor where it is OK. Last edited by lua; April 19th, 2018 at 11:51 PM.  
April 20th, 2018, 05:29 AM  #2 
Newbie Joined: Sep 2013 From: Reno,NV Posts: 14 Thanks: 1 Math Focus: Number Theory 
90^{\circ}=$90^{\circ}$

April 21st, 2018, 10:20 PM  #3 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
Can anybody help?

April 24th, 2018, 08:21 AM  #4 
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140 
Draw $\displaystyle AX \parallel BC$ From the parallel lines we have $\displaystyle \angle AXE = \angle CDG = \angle CPQ $ and $\displaystyle \angle AEX = \angle CQP$. But $\displaystyle \angle CQP = \angle CPQ$ so $\displaystyle \angle AEX = \angle AXE$ and we have that $\displaystyle AE = AX$. Now observe that $\displaystyle \triangle AXG \cong \triangle BDG$ so we have that $\displaystyle AX = BD$. Thus $\displaystyle AE = BD$ 
April 24th, 2018, 12:19 PM  #5 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
Thanks.

April 25th, 2018, 07:01 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
Because I am so picky, I feel the need to point out that the two line segments are NOT "equal" as implied in the title, They are "congruent" because they have the same length: AE= BD, as stated in the problem, where AE and BD are the lengths of line segments $\displaystyle \overline{AE}$ and $\displaystyle \overline{BD}$. In mathematics to say that $\displaystyle \overline{AE}$ and $\displaystyle \overline{BD}$ are equal would mean that $\displaystyle \overline{AE}$ and $\displaystyle \overline{BD}$ are different names for the same line segment. 
April 25th, 2018, 09:14 AM  #7 
Senior Member Joined: Oct 2009 Posts: 771 Thanks: 278  
April 25th, 2018, 11:44 AM  #8  
Senior Member Joined: Feb 2010 Posts: 706 Thanks: 140  Quote:
 
May 5th, 2018, 11:05 AM  #9 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
I am! I just thought that bitching about the line segments was enough!


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