Let BAC be the give right angle.
- With centre A and any convenient radius , draw an arc cutting AB at Q and AC at P
- With the same radius and centres P and Q , draw arcs cutting the arc of step a) at R and S respectively.
- Draw lines joining A with P and Q. AP and AQ trisect the right angle BAC.
Thus ∠BAR = ∠RAS = ∠SAC = 1/3 ∠BAC
Let AB be the given line segment to be divided into unequal parts say 1/6 ,1/5 ,1/ 4 ,1/3 ,1/ 2 .
- Draw a line segment AB of given length.
- Draw perpendiculars AD and BC at points A and B. Complete the rectangle ABCD
- Join diagonals AC and BD intersecting at E.
- Draw perpendicular from E on AB and find intersection point F.
- Now AF = 1/ 2 AB
- Join D and F. The segment FD intersects the diagonal AC at G. Drop perpendicular from G to A. Then AH = 1/3 AB
- Similarly make constructions as given in the figure to obtain 1/ 4 AB , 1/5 AB and 1/6 AB
Following are the steps for construction of a perpendicular to a line from a point outside it. In this construction, the arcs are shown as circles.
- Let AB be the given line and P is the point outside it
- With A as centre and radius AP, draw an arc cutting the given line at C.
- With C as centre and radius CP, draw an arc cutting the arc of step 2 at R.
- Draw line segment joining PR intersecting AB at Q.
PQ is the required perpendicular. Drag Point P and see the effect.
