Dividing a Line Segment
Date: 03/08/2001 at 02:14:08 From: Ekmel Subject: Geometry - angle bisector How do I divide a line segment into a 2:3 ratio using the angle bisector? So far: draw auxilary line construct parallel line choose a random compass setting and mark it off
Date: 03/08/2001 at 15:28:20 From: Doctor Rob Subject: Re: Geometry - angle bisector Thanks for writing to Ask Dr. Math, Ekmel. Here is the diagram: C / P2 o / \ / \ P1 o \ / \ / \ / A o-----------o-----------------o B / X \ / \ / \ o Q1 \ / \ / \ o Q2 \ / \ / o Q3 / D Through A draw an auxiliary line AC. Through B draw a line BD parallel to AC. Set the compass at some opening, and from A mark P1 on AC. From P1 mark P2 on AC. From B, on the opposite side of AB mark Q1 on BD. From Q1 mark Q2 on BD. From Q2 mark Q3 on BD. Now connect P2Q3. The point where this intersects AB is X, which satisfies AX:XB = 2:3. I leave it to you to prove that this works. Hint: triangles AXP2 and BXQ3 are similar. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
Date: 03/11/2001 at 18:17:34 From: The Lemke's Subject: Re: Geometry - angle bisector Thanks. I had gotten just about that far on my own - but I still do not see how the angle bisector fits into it. Any more hints?
Date: 03/12/2001 at 08:32:21 From: Doctor Rob Subject: Re: Geometry - angle bisector Thanks for writing back! Here's another way that does use an angle bisector: Given line segment AB, mark off from A along AB points P1, P2, and P3 such that AP1 = P1P2 = P2P3. With radius AP2, draw a circle with center A. With radius AP3, draw a circle with center B. Let one of the points of intersection of the two circles be C. Connect AC and BC. Now bisect <ACB. Let X be the intersection of that bisector and AB. Then AX:XB = 2:3. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/
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