Proof Using Coordinate GeometryDate: 06/08/2001 at 13:10:19 From: Andrew Subject: Hughmoar County Law #98 I have a problem from my geometry class that I can't seem to figure out. Basically, the problem is that in equilateral triangle ABC, with side length 9 miles, a segment is drawn from point A to side BC at a point (D) 3 miles from point C, and a segment is drawn in a similar manner from point B to a point (E) on side AC so that it is 3 miles from point A. At the intersection of these segments is point F, from which a segment is drawn to point C. The problem is to try to prove that angle BFC (or EFC) is a right angle. I just can't figure out how to do this. Thanks! Andrew Date: 06/11/2001 at 08:33:39 From: Doctor Rick Subject: Re: "Hughmoar County Law #98" Hi, Andrew. Let's draw a picture: This doesn't seem like a very simple geometry problem. I thought about it over the weekend, but I couldn't prove the result using classical geometry. That's not to say it can't be done; if you have a proof by now, I'd like to see it. While exploring the geometry, I did discover one interesting and possibly helpful fact: angle BFD has measure 60 degrees. I proved this using the fact that ABC is equilateral, and the fact that AE equals CD - but NOT using the fact that AE = AC/3. Thus it is true even if AE and CD are some other length than AC/3, as long as they are equal. The proof of this fact involves the congruence of triangles ABE and CAD, and the fact that angle AFE is an exterior angle of triangle ABF. I assume that angle DFC is 30 degrees only when AE = AC/3, but I have not yet found a way that I could make use of the length of AE - using classical geometry, that is. On the other hand, a proof using coordinate geometry is quite straightforward. I chose a coordinate system in which C is at the origin (0,0) and A is at (1,0) (I have divided the lengths by 9 feet for convenience). Then B is at (1/2, sqrt(3)/2). You can work out the coordinates for D and E, for instance, E is 2/3 of the way from the origin (C) to (1,0) (A), so it is at (2/3, 0). Next, you can write equations for the lines AD and BE. Solve the equations simultaneously to find the coordinates of F. We're almost done. Just find the slopes of CF and BF, and use them to show that the lines are perpendicular. By the way, I can't figure out the subject line of your question. Was the problem posed in terms of a fictitious or real zoning law, or what? - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ Date: 07/09/2003 at 04:56:55 From: Aaron Subject: Alternate (geometric) solution I discovered a geometric solution to the problem. In your reply to Andrew, you state that angle BFD = 60 deg Then angle EFD = 120 deg Because angle ECD = 60 deg quadrilateral ECDF is cyclic (i.e., pairs of opposite angles sum to 180 degrees). Then angle CFD = angle CED But angle ECD = 60 deg and EC/CD = 2 So ECD is a 30-60-90 triangle, so angle CFD = angle CED = 30 deg and angle CFB = angle CFD + angle DFB = 30 deg + 60 deg = 90 deg Date: 07/09/2003 at 13:11:46 From: Doctor Rick Subject: Re: Alternate (geometric) solution Hi, Aaron. Good work! The observation that CDFE is a cyclic quadrilateral really opens up the problem. Once we have determined in addition that ED is perpendicular to CD, we can go directly to the conclusion, since angle CDE = angle CFE. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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