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Proof Using Coordinate Geometry

Date: 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!


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 

- Doctor Rick, The Math Forum   

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


  angle EFD = 120 deg


  angle ECD = 60 deg

quadrilateral ECDF is cyclic (i.e., pairs of opposite angles sum 
to 180 degrees).  Then 

  angle CFD = angle CED


  angle ECD = 60 deg


  EC/CD = 2

So ECD is a 30-60-90 triangle, so 

  angle CFD = angle CED 

            = 30 deg


  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   
Associated Topics:
High School Coordinate Plane Geometry
High School Geometry
High School Triangles and Other Polygons

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