Geometry Forum - Problem of the Week
Solutions - Construction - Angle, April 11-15, 1994
Eric Chan
____________________
A /
/
/
X /
/ P
Q /_______
/
O _________________ B
Coonstruct PQ // OB Q on OA
Construct X on OA such that OQ = QX
Draw XY to intersect OB at Y
XY is the line.
Daniel Chan
____________________
A /
/
/
X /
/ P
C /_______
/ /
O _______/___________ B
D
Draw a line through P parallel to OB meeting OA at C. Then draw
another line through P parallel to OA meeting OB at D, then draw
CD. Finally draw a line through P parallel to CD meeting OA at X
and OB at Y. P is the midpoint of XY.
Daniel Chan & Scott Greenfield
____________________
A /
/
X /
-----------------
/ /
C / P/
/ /
O _______/___________ B
Y
Using the Geometer's SketchPad we draw angle AOB and a
point P inside the angle. Then we chose P as a center of
rotation. Rotate AOB about 180 degrees about P. Then we draw
a line XY between the points of intersection of the original
drawing and its image. P is the midpoint of this line.
Patson Settachutgal
____________________
Construct a perpendicular from P to ray OB. The point where these intersect will be point Z. Construct a line through P that is parallel to ray OB. The point where these intersect will be point D. Construct another line parallel to line PD that intersects ray OA. (The distance from this line to line PD equals the length of segment PZ-- call this equal distance PE). The point where this line intersects ray OA is point X. Construct the line from point X through point P to ray OB. The point where this line intersects ray OB is point Y. P is the midpoint of segment XY because triangle YPZ is congruent to triangle XPE.
Meera Shah
____________________
Draw angle AOB with point P inside of it. Draw the perpendicular from P to ray OB. Call the point on ray OB Z. Draw a line through P parallel to ray OB and perpendicular to PZ. Extend the line through ray OA. Take compass of radius PZ. Construct a circle w ith center P and radius PZ. Call the point where the circle and line PZ intersect near ray OA M. Draw line MX parallel to ray OB where X is on ray OA. Draw a line XP through P to intersect ray OB. Call the intersection Y. P is the midpoint of segment XY because PX = PY through ASA (Angle-Side-Angle). Angle XMP and angle YZP are equal because alternate interior angles determined by two parallel lines and a transversal are congruent. MP = PZ because they are both radii of the same circle. Angle MPX and angle YPZ are congruent because vertical angles are congruent. The triangles are congruent, thus PX = PY and P is the midpoint.
Gino Perotte
____________________
I found a way to construct the correct segment every time you
try to do the problem. My solution is independent of the type of
angle, or where the dot is in the interior. This is how to construct a
segment such that the point in the interior of the triangle is the
midpoint of the segment.
First, through the point in the interior, draw the two lines that
are parallel to the sides of the angles. This forms a parallelogram.
Then draw the diagonal of the parallelogram that does not connect
the interior point with the vertex. This is the distance that each half
of the segment must be. So now, draw a segment with the diagonal
distance from the interior point to the two sides of the angle. All
three points are collinear, and the distance from the intersection of
each side of the angle to the interior point is equal, thus making the
interior point the midpoint of the total segment.
angle ABC:
| /
| /
| /
| A ____________.p
| / /
| / /
| / /
-----B------------m---------------C---------------------------------
|
|
Segment Am is the diagonal to be useed. Use a segment of
equal length to Am and have it begin at point p and end at it's point
of intersection with segment BA. Likewise, use a segment of equal
length to Am and have it begin at point p and have it end at it's point
of intersection with segment BC. The two segments have equal
length and all three points (point p, the intersection of BA, and the
intersection of BC) are collinear.
You can prove this by plotting the angle on a graph and using
(0,0) as the coordinates for the angle vertex. Then by using the
pythagorean theorem, you can find the length of the diagonal and
then draw in the points. Then you can test the slopes of the two
segments formed to see if they have the same slope. If they do, then
they are collinear. And then you can use the pythagorean theorem
to see is each segment is congruent to eachother. If they are, then
point p is the midpoint of the entire segment.
Hannah Guhm
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Given angle AOB and point P in the interior, construct a parallel line through P parallel to ray OB, intersecting ray OA at D. Construct a segment DX on OA such that DX equals OD. Therefore, transversal OX is divided into segments in the ratio 1:1. Draw XP until it intersects OB at Y. This transversal is also divided in the ratio 1:1 and so P is its midpoint.
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