Construct segment EF - February 19-23, 1996
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Here's a bit of a construction problem for you, though not the sort of construction we've been talking about recently. You've been sending in so many solutions recently, I thought I should give you a challenge.Given an angle ABC, and any point D in the interior of angle ABC, construct a segment EF such that D is the midpoint of segment EF, E lies on ray BA, and F lies on ray BC.
Annie says: Solutions
We stumped a bunch of you this week, and we can see from the solutions where constructions are dealt with carefully in the curriculum - and it isn't on my continent! What a shame... they're so much fun. :-)
Almost all of the incorrect submissions involved the same mistake - they found an E and an F and then found D as the midpoint of EF. Basically they tackled it backwards. This makes it a much less interesting problem, as there isn't any challenge at all. You really need to read the problem carefully - Given A, B, C, and D, figure out how to find E and F. A neat construction problem is like a puzzle that will take a little bit of thought before the solution reveals itself to you.
There are several ways to get something that works here. The students from Loreto College submitted the most 'formal' solutions, as they even cited a theorem., but you'll find several different approaches included, including graphics folks sent with their solutions. Following are highlights. The names of all the people who submitted correct solutions and most of the solutions are also available.
Therese Quinn Year 10 Loreto College, Marryatville Adelaide, South Australia. THE MIDPOINT THEOREM The line joining the midpoint of two sides of a triangle is parallel to one half the length of the third side. THE CONVERSE OF THE MIDPOINT THEOREM The line through the midpoint of one side of a triangle parallel to the base always bisects the third side.In an angle ABC, with a point D anywhere in the interior of the angle, to construct a segment EF such that D is the midpoint of segment EF, where E lies on ray BA and F on ray BC, I used the converse of the midpoint theorem above. In creating segment EF, a triangle is created. D is the midpoint of one of the sides of the triangle. If a line parallel to the base, where ray BC is the base, is drawn through D, according to the converse of the midpoint theory, the intersection of ray BA and the line would be the midpoint of another side of the triangle created by constructing segment EF. Using ray BA as the base instead of ray BC, a second parallel line can be drawn, and a second intersection formed. So, with point D, and the two intersections, the three midpoints of a triangle are formed. The vertex of angle ABC is one corner of the triangle. If two symmetrical lines are constructed extending the distance between B and the two intersections to make a point twice as far away as B on each ray, points E and F are created, the other two corners of the triangle. If E and F are joined, they create a line with D as its midpoint. If this construction is made using Cabri Geometry, point D can be moved, causing the parallel lines and intersections to move, thus making the segment EF move, keeping D as its midpoint.
Caitlin Gill Year 10 Loreto College, Marryatville Adelaide, South Australia. A. . . . E . . . . . . . . . . . J .___________.___________________ . D . . . . . . . . . B ._______________________.________________C F Given the angle ABC and that point D is any point in the interior of the angle, show that for any D we can construct EF such that D is the midpoint of EF and E is on BA and F is on BC. We first draw JD parallel to BC with J on BA. Now let E be the point on BA such that BJ equals JE. Now extend the line ED to meet BC in F. We will show that our construction has produced the required segment EF. Our proof is based on the "Converse of the Midpoint Theorem". STATEMENT: "The line drawn from the midpoint of one side of a triangle, parallel to a second side, bisects the third side". PROOF: Given triangle BEF has J as the midpoint of side BE and JD is parallel to BC. We seek to show that D = midpoint of EF. Now triangle BEF is similar to triangle JED (equiangular triangles) since 1) Angle E is common 2) Angle EJD is equal to angle EBF ( corresponding angles in parallel lines, as JD is parallel to BF). 3) Angle EDJ is equal to angle EFB ( corresponding angles in parallel lines, as JD is parallel to BF). Therefore the sides of triangle BEF have a common ratio to the corresponding sides of the similar triangle JED. And hence, as EJ is half of BE, it follows that ED is half of EF. So D is the midpoint of EF.
Thomas S. Kuo Grade: 7 School: Murray Junior High School, Ridgecrest, California A * * E * * * * * * * * D * * * * * * * * * * * B * * * * * * * * * * * * * * C G F Make a line DG through point D and parallel to AB. Point G lies on ray BC. Locate point F on ray BC such that GF = BG. Draw a line through point F and D and intersects BA at point E. Segment EF is what we need. I know triangle FDG and triangle FEB are similar because DG is parallel to AB. I also know that FD/FE = 1/2 because FD = DE. Then FD/FE = 1/2 = FG/FB. Therefore BG = GF and this leads me to get the idea to construct EF. Similarly, since I know DG/EB = FD/FE = 1/2. I can locate point E on ray BA such that BE = 2 * GD. I then connect point E and D and intersects ray BC at F and construct segment EF.
Ken Duisenberg M.S. Electrical Engineering Hewlett-Packard Let Dy be the perpendicular distance from ray BC to point D. Then the problem consists of finding point E on AB a distance 2*Dy from ray BC. Then, extending DE to F on BC will give the desired segment. Here is one way: Construct circle O, with center D, to intersect BC at G and H. Draw lines GD and HD to intersect circle O at points I and J. Draw line IJ to intersect ray AB at point E. Draw line ED to intersect BC at point F. To see that this works, examine the two right triangles, one with height 2*Dy (point E) and one with height Dy (point D), and both with the other acute angle at point F. The similar triangles show that D must be the midpoint of EF. An attempt at a drawing: A / / Circle O / --- ____/____/___\___________ /E /J I\ / ( . ) / \ D / B /________\___/_____F___C G --- H
Brian Gordon Dartmouth '92 I thought this was going to be ridiculously difficult, but after looking at a special case of the problem, it then got extremely easy. Step 1. Construct lines through D parallel to both rays BA and BC. This creates a parallelogram with B and D at opposite corners. Step 2. Construct two more congruent parallelograms. Both will be contained inside angle ABC, one on each of the sides of the first constructed parallelogram. It should look something like this: A \ \-------\ \-------\D------\ B\-------\-------\------------C It should appear that the parallelogram has been used in a tessellation inside the angle. If we draw the diagonals for the two most recently constructed parallelograms, they will both contain D. They will be equal, as they are from congruent parallelograms. And best of all, they are concurrent, as the angles formed with the sides of the parallelograms are congruent. And the endpoints are on the sides of angle ABC, so all the requirements are met. I was surprised how easy it got when I looked at the case of ABC being a right angle. It was easy to look at a pattern of rectangles, but the rules for oblique parallelograms still meet the criteria of the problem. --bri
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