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Triangle Proof


Date: Sat, 18 Feb 1995 22:01:29 -0800 (PST)
From: Ralph Sierra
Subject: Help, please!

I'm a student in high school and I need help on this one problem.
I'm in geometry and stumped. I'm thinking that maybe to figure it 
out, it needs to be said that if 2 sides of a triangle are not 
congruent, then the angles opposite them are not congruent, and the 
larger angle is opposite the longer side. But I'm not sure how to 
say it in a proof. I have an idea but am getting a little confused. 
If you could help me, I would appreciate it.  

                         A
                        /|\
                       / | \
                      /  |  \
                     /   |   \
                    /    |    \
                   /     |     \
                  /      |      \  
                 /       |       \ 
                /        |        \
               /_________|_________\
             B           D           C   

    B, D and C are points of the line L such that B-D-C and BD<DC.
    If segment AD is perpendicular to L, prove that AB<AC.

Thank you,
- Michelle


Date: Sun, 19 Feb 1995 02:16:52 -0500 (EST)
From: Dr. Ken
Subject: Re: Help, please!

This is a job for the law of Sines.  As you may know, if A, B, 
and C are the angles of a triangle and a, b, and c are the sides 
opposite them, then 

          a         b       c
        ------ = ------ = ------
        Sin[A]   Sin[B]   Sin[C].

So in your case, if what you need to show is that AB<AC, 
a good way to do that would be to show that angle C is smaller 
than angle B. Then you'd be able to say that 

  AB          AC
-------- = --------
 Sin[C]     Sin[B], 

And since Sin[C] < Sin[B], AB<AC.  Does this help you?  
If you buy that, then all you have to do is show that angle C 
is smaller than angle B. I hope this train of thought gets you 
somewhere, and write back if you're still stumped.

-Ken "Dr." Math


Date: Sun, 19 Feb 1995 08:50:45 -0800 (PST)
From: Ralph Sierra
Subject: Re: Help, please!

In our class we have not studied the law of sines yet, so 
I'm not following your directions.  I'm sure you're correct, 
but can you please explain it a little simpler?  Thanks 

- Michelle


Date: Sun, 19 Feb 1995 12:57:34 -0500 (EST)
From: Dr. Ken
Subject: Re: Help, please!

Hello there.

Not allowed to use the law of sines, huh?  Well, perhaps we can 
do this with the Pythagorean theorem.  Oh, wow, in fact it's 
easiest this way.  

We know that BD<CD.  Using the Pythagorean theorem 
(BD^2 + AD^2 = AB^2 and CD^2 + AD^2 = AC^2), what can you conclude 
about the lengths of AC and AB?  I think this is an easier way 
than the way I was trying to explain before.

-Ken "Dr." Math
    
Associated Topics:
High School Geometry
High School Triangles and Other Polygons

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