Tends to infinity

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Another Geometrical construction…

Posted by tendstoinfinity on February 5, 2011


I in previous post posed a geometric problem

An angle and a point D lying inside it are given. Construct a line segment with end points on arms of the angle and bisected at D.

let us consider a set of points on a line  BA  and all the segments having one end on these points and mid point at D. then the other end of each line segment will be just the reflection of  the former points. hence the other ends will always lie on a line parallel to BA.

then the above construction can be easily done.

suppose the given angle is <APQ. Image of the vertex P of the angle with respect to D is drawn (draw a line through P and D. cut an arc of length equal to PD from centre D. where it cuts the line is image P’ of P with respect to D

draw a line parallel to PA through P’ . suppose this parallel cuts PQ at R then RS will be required line which has mid- point at D and ends on the arms of the angle

surely you can justify this. I will give a very good  use of this construction

An angle \angle APQ and a point D lying inside it are given. Draw a straight line through D which makes a triangle with the arms of the angle of minimum possible area.

Then obviously the triangle RPS (with reference to the above figure) will have minimum area amongst all the triangles of this type !! got it.


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