November 4th, 2014, 08:30 PM  #1 
Newbie Joined: Nov 2014 From: Ohio Posts: 3 Thanks: 0  Congruent Triangles within a parallelogram http://dmmalkin.ipage.com GIVEN: AC is parallel to BD AC is congruent to BD CF is congruent to EB PROVE: Use 2 columns Triangle AFC = Triangle BED 
November 5th, 2014, 06:24 AM  #2 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
I think it is not true (see the attached image). AC is parallel to BD AC is congruent to BD CF is congruent to EB but triangle AFC is not equal to triangle BED. Last edited by skaa; November 5th, 2014 at 06:51 AM. 
November 5th, 2014, 08:43 AM  #3 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 
should be $\displaystyle \Delta AFC \cong \Delta DEB$ use alternate interior angles to show $\displaystyle \angle CAE \cong \angle BDF$ , then using the two congruent sides, the two triangles are congruent by AAS 
November 5th, 2014, 10:29 AM  #4 
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77 
Two congruent sides can be like SSA but not AAS that is two congruent angles (angleangleside) .
Last edited by skaa; November 5th, 2014 at 10:35 AM. 
November 5th, 2014, 12:39 PM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1588 
Yes, my mistake ... saw two pairs of congruent sides and thought two congruent angles. I still think there is a way to show the two triangles are congruent. Have to think on it. 
November 5th, 2014, 07:17 PM  #6  
Newbie Joined: Nov 2014 From: Ohio Posts: 3 Thanks: 0  Quote:
Intuitively, we can deduce that this makes these 2 lines parallel See my attachment I know that my logic is correct but I don't know how to show this as a formal 2 column proof Click here Last edited by dmmalkin; November 5th, 2014 at 07:21 PM.  
November 6th, 2014, 12:28 AM  #7 
Senior Member Joined: Jul 2014 From: à¤à¤¾à¤°à¤¤ Posts: 1,178 Thanks: 230  The diagram attached by skaa says that $\displaystyle \textrm{CF} \cong \textrm{EB}$. But it is not. Have a look at the real diagram : $ \usepackage{color} \setlength{\unitlength}{2144sp}% % \begingroup\makeatletter\ifx\SetFigFont\undefined% \gdef\SetFigFont#1#2#3#4#5{% \reset@font\fontsize{#1}{#2pt}% \fontfamily{#3}\fontseries{#4}\fontshape{#5}% \selectfont}% \fi\endgroup% \begin{picture}(12450,4260)(751,4621) \thicklines {\color[rgb]{0,0,0}\put(4636,691){\line(1,1){3645}} \put(991,4336){\line( 1, 0){8415}} \put(9406,4336){\line( 1, 1){3645}} \put(13051,691){\line(1, 0){8415}} }% {\color[rgb]{0,0,0}\put(2701,2356){\line( 1,1){315}} }% {\color[rgb]{0,0,0}\put(2521,2491){\line( 1,1){315}} }% {\color[rgb]{0,0,0}\put(1036,4336){\line( 5, 1){6654.808}} }% {\color[rgb]{0,0,0}\put(13051,691){\line(6,1){6808.378}} }% {\color[rgb]{0,0,0}\put(4681,691){\line( 4,3){4773.600}} }% {\color[rgb]{0,0,0}\put(11071,2401){\line( 1,1){315}} }% {\color[rgb]{0,0,0}\put(11251,2221){\line( 1,1){315}} }% {\color[rgb]{0,0,0}\multiput(4591,3436)(2.54717,7.64151){54}{\makebox(6.3500,9.5250){\small}} }% {\color[rgb]{0,0,0}\multiput(9361,1141)(2.54717,7.64151){54}{\makebox(6.3500,9.5250){\small}} }% \put(4501,511){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefa ult}{\updefault}{\color[rgb]{0,0,0}A}% }}}} \put(13186,511){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefa ult}{\updefault}{\color[rgb]{0,0,0}B}% }}}} \put(766,4606){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefa ult}{\updefault}{\color[rgb]{0,0,0}C}% }}}} \put(9586,4606){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefa ult}{\updefault}{\color[rgb]{0,0,0}D}% }}}} \put(6076,2086){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefa ult}{\updefault}{\color[rgb]{0,0,0}E}% }}}} \put(7876,3031){\makebox(0,0)[lb]{\smash{{\SetFigFont{12}{14.4}{\rmdefault}{\mddefa ult}{\updefault}{\color[rgb]{0,0,0}F}% }}}} \end{picture}% $ It is given that $\boxed{\\ \ \\ \textrm{AC} \parallel \textrm{BD} \\ \ \\ \textrm{AC} \cong \textrm{BD} \\ \ \\ \textrm{CF} \cong \textrm{EB} \\ }$ So, as in skaa's diagram, if point $\displaystyle \textrm{F}$ is moved, it will not be $\displaystyle \cong \textrm{EB}$ unless it is moved at the mirror image of its initial position. I think that the question is easy. According to the given condition, $\angle \textrm{ACF}$ has to be = $\angle \textrm{BED}$ So, $\textrm{SAS}$ congruence criterion. There may be other methods. We need to prove that $\bigtriangleup \textrm{AFC}= \ \bigtriangleup \textrm{BED}$. But equality does not mean congruency. So, maybe, we need to prove their areas to be equal. Last edited by Prakhar; November 6th, 2014 at 01:19 AM. 
November 6th, 2014, 06:21 AM  #8  
Senior Member Joined: Mar 2011 From: Chicago, IL Posts: 214 Thanks: 77  Quote:
Hence, BE=CF. So, my logic is correct, too. Last edited by skaa; November 6th, 2014 at 06:24 AM.  
November 6th, 2014, 06:28 AM  #9  
Newbie Joined: Nov 2014 From: Ohio Posts: 3 Thanks: 0  Quote:
Logically, if you move F, you MUST move E BY THE SAME DISTANCE in order to keep the given statement of "CF = EB" true. (also, there is no grounds to just assume that AFC is congruent to BED) I just need help with the formal proof based on logic shown by me above. Last edited by dmmalkin; November 6th, 2014 at 06:32 AM.  
November 6th, 2014, 07:28 AM  #10  
Global Moderator Joined: Dec 2006 Posts: 20,931 Thanks: 2205  Quote:
In the original diagram, points A, E, F and D appear in that order on the line segment AD, but the notes don't require that order. In skaa's diagram, the order is changed to A, F, E and D, which allows the falsity of what has to be proved. Thus, the original problem is flawed and a proof cannot be provided.  

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