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alexb
Charter Member
1696 posts |
May-17-05, 01:11 PM (EST) |
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1. "RE: A Parallel Question"
In response to message #0
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Let's look at this from a little different angle. Start with your triangle ABC and form an equilateral triangle ACE away from B. Then count angles. In order to do that, introduce a point B' on AB so that A is inside the segment B'B. So for the angles we have B'AE + EAC + CAB = 180 Since EAC = 60 (1) B'AE + CAB = 120 In triangle ABC, CAB + ACB + ABC = 180 Since ACB = 60, (2) CAB + ABC = 120. From (1) and (2), B'AE = ABC, such that AE||BC. If BD = AC, then the quadrilateral AEDB is a parallelogram; and ED||AB. |
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Deep G
Member since Nov-6-05
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Nov-19-05, 10:36 AM (EST) |
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3. "RE: A Parallel Question - Use contradiction !"
In response to message #1
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Hi, Alex started with AE||BC (Constructing an equilateral tri. AEC implies construction of two alternate angles ACB and CAE of the same measure 60 degrees.i.e. in effect constructing AE parallel to BC) and showed it to lie on line m which is parallel to AB (i.e. showing AEDB a parallelogram) Here's a different approach. Suppose AE is not parallel to BC where EC=AC and E lies on line m as given in the problem. Construct AE' parallel to BC (i.e. BD too) and let it intersect line m at E'. Note that E' lies on the same side of AC as E lies. consider quad. BDE'A. AE' parallel to BD and AB parallel to DE' (i.e. line m as given). So its a parallelogram. So BD=AE'. But BD=AC.So AE'=AC .........................(1) Since we constructed AE' parallel to BC, angle CAE' = angle ACB = 60 degrees.........................(2) (1) and (2) imply that ACE' is equilateral - meaning AC=CE'.
That means there are two distinct points E, E' on line m at the same distance (equal to AC) from point C and lying on the same side of AC.
CONTRADICTION ! So, our assumption that AE is not parallel to BC is false. So AE parallel to BC. Any comments? :) |
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JJ
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Nov-22-05, 09:58 AM (EST) |
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4. "RE: A Parallel Question"
In response to message #0
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Suppose AE parallel to BC :
Then ABDE would be a parallelogram,
in consequence AE=DB=AC, so triangle ACE would be equilateral,
then, Angle(ECA) = Angle(CAB) = 60°
in triangle ABC, Angle(ACB) = 60° and Angle(CAB) = 60°,
so, triangle ABC would be equilateral, which is contradictory with BC>AB.
So, the supposition is false and AE isn't parallel to BC.
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Deep G
Member since Nov-6-05
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Nov-24-05, 11:03 AM (EST) |
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5. "RE: A Parallel Question"
In response to message #4
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>Suppose AE parallel to BC :
>Then ABDE would be a parallelogram,
>in consequence AE=DB=AC, so triangle ACE would be
>equilateral, So far so good. No 2 ways about it. >then, Angle(ECA) = Angle(CAB) = 60° Angle(CAB)=60 degrees ??? Is this given in the problem? Or can it be concluded from arguments you put forward earlier? Either I am missing something or you are ! |
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JJ
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Nov-27-05, 07:39 AM (EST) |
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6. "RE: A Parallel Question"
In response to message #5
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Sorry, I made a writing mistake. The correct relationship is :
Angle(ECA) = Angle(CAD) = 60° and triangle CAD is equilateral.
CEAD is a lozenge. ED and AC are perpendicular.
Angle(CED)=Angle(CDE)=Angle(AED)=angle(ADE)=30°
AB parrallel to ED and AC perpendicular implies angle(BAC)=90°
In fact, CE=CA=CD=BD. Hence D is midpoint of BC.
In conclusion, if in triangle ABC we have angle(BAC)=90°, angle(ACB)=60°, Angle(CBA)=30°, then EA is parralel to BC. |
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Deep G
Member since Nov-6-05
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Nov-29-05, 07:34 AM (EST) |
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7. "RE: A Parallel Question"
In response to message #6
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>Sorry, I made a writing mistake. The correct relationship is
>:
>Angle(ECA) = Angle(CAD) = 60° and triangle CAD is
>equilateral.
>CEAD is a lozenge. ED and AC are perpendicular.
>Angle(CED)=Angle(CDE)=Angle(AED)=angle(ADE)=30°
>AB parrallel to ED and AC perpendicular implies
>angle(BAC)=90°
>In fact, CE=CA=CD=BD. Hence D is midpoint of BC.
>In conclusion, if in triangle ABC we have angle(BAC)=90°,
>angle(ACB)=60°, Angle(CBA)=30°, then EA is parralel to BC. I am afraid u have unknowingly assumed certain things in while providing your solution. As I am having difficulty in understanding how you got various constraints that ought to be there for AE to be arallel to BC, I request you to once go through my solution and point out inadequacies or incorrect steps, if any. My solution is a generic one and states that AE||BC irrespective of D being or not being the mid-point of BC. : )
Regards,
Deep G |
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