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November 4th, 2014, 08:30 PM   #1
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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
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November 5th, 2014, 06:24 AM   #2
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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.
Attached Images
File Type: jpg Forum.jpg (10.9 KB, 11 views)

Last edited by skaa; November 5th, 2014 at 06:51 AM.
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November 5th, 2014, 08:43 AM   #3
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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
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November 5th, 2014, 10:29 AM   #4
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Two congruent sides can be like SSA but not AAS that is two congruent angles (angle-angle-side) .

Last edited by skaa; November 5th, 2014 at 10:35 AM.
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November 5th, 2014, 12:39 PM   #5
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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.
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November 5th, 2014, 07:17 PM   #6
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Quote:
Originally Posted by skaa View Post
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.
Don't forget that BE = CF (essentially, these 2 line segments move away from the center point at the same rate).

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.
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November 6th, 2014, 12:28 AM   #7
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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.
Thanks from Ubuntu

Last edited by Prakhar; November 6th, 2014 at 01:19 AM.
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November 6th, 2014, 06:21 AM   #8
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Quote:
Originally Posted by dmmalkin View Post
Don't forget that BE = CF (essentially, these 2 line segments move away from the center point at the same rate).

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
In my diagram




Hence, BE=CF. So, my logic is correct, too.

Last edited by skaa; November 6th, 2014 at 06:24 AM.
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November 6th, 2014, 06:28 AM   #9
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Quote:
Originally Posted by prakhar View Post
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.
Unfortunately, your picture just shows as code.
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.
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November 6th, 2014, 07:28 AM   #10
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Quote:
Originally Posted by prakhar View Post
The diagram attached by skaa says that $\displaystyle \textrm{CF} \cong \textrm{EB}$. But it is not.
On the contrary, it is, and in skaa's diagram what has to be proved isn't true.

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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