How to Drop a Perpendicular from a Given Point to a Given Straight Line.

To drop a Perpendicular from a Given Point to a Given Straight Line.

  Let it be required to lay out a line perpendicular to the line AB

from a given point C (Fig. 20). Fasten the end of the chain at C, or get a chainman to hold it there, and choose some convenient length CD to form one side of an isosceles triangle CDE. Get an assistant to hold the graduation mark on the chain at D and, standing at G a short distance behind A, signal

to him to move the tightened chain until D appears to be in line with AB and get him to put in an arrow. Have a similar arrow put in at E, where CE = CD and E is on the line AB. Measure the distance DE and put in an arrow at F on the line AB such that DF = ½DE. The line FC is then at right angles to AB and passes through C.

  

  If the point C is inaccessible, as in Fig. 21, choose suitable points D and E on the line AB and from D and E lay out lines DF and EG perpendicular to EC and DC respectively.

  Then line out a point H so that it lies on the intersection of the lines DF and EG. Finally, from H lay out HK perpendicular to AB. The line KH when produced should then pass through C, as should be verified by standing

behind a pole at K and seeing if K, H and C are on one straight line.