Hints for construction tasks

Obtain hints to relate non-given objects to the given ones by construction.

Construction tasks ask for construction steps (equivalent to using a compass and ruler (or a square set )) that lead from given objects to the object to be constructed. Dynamic geometry systems can help solve geometric tasks in a variety of ways, ranging from the study of loci of points to visual verification of constructions by dragging points.

We show here a method specific to OK Geometry that often gives not only useful hints for construction, but also a pleiad of several, often unexpected, solutions to construction tasks.

Here is the general approach:
  1. Obtain, in some way, a configuration with the given objects and the object(s) to be constructed. You can start from the final object(s) and then position accordingly the given objects. Another option is to use advanced construction commands in OK Geometry.
  2. Use OK Geometry to observe the configuration.
  3. Choose the observed the properties that relate the object(s) to be constructed to the given ones. At this stage, it is of great help to use the query buttons (magenta buttons with question marks on the right) - to show only properties related to a specified object.
  4. Elaborate the construction steps.
Notes.
  1. If no satisfactory property is found, it often helps to add to the configuration new points or objects - constructible from the given ones.
  2. The observed properties used in the construction should, obviously, be proved to validate the construction.
  3. The Plus mode of OK Geometry contains a much more powerful mechanism for relating constructed objects to the initially given ones. See this link.
  4. The Plus mode also includes a method to display the steps of geometric construction as an animated image or a sequence of images. See link.
We illustrate the method with a simple example.


Given is a line and points A, B on the same side of k
How to construct a circle through A and B that touches the line ?

In general, there are, obviously, two different solutions to the task.

First, you need a configuration to be observed. You can draw the given objects (points A,B and line k) and use the command Circle 3obj to position the sought circle. Or, you start with a circle and position the points A, B and the line k.

The obvious strategy in solving this construction task is to find the position of the centre of the sought circle  or a third point on its circumference - the obvious chouce is the point ov contact of the circle and the line k. In the above figure (with eventually added objects) OK Geometry detects properties that lead to different solutions of the task above. We show some properties and the related constructions. Proving some of them can be challenging task.

In the figures below we use the following notation:
D and D' are the points of contact of the sought circles with the line k,
C is the intersection line AB and line k,
F is the intersection of and the bisector of AB,
G and G' are the centres of the sought circles,
E is the midpoint of GG'.

Triangles ACD and DCB are similar.
Consequently, |CD|2 = |CA|·|CB| .
 
The construction is based on a well known property of right triangles.
 
Steps:
  1. Let C be the intersection of line AB and k.

  2. Draw a semicircle kBC with BC as diameter.

  3. Let P be the intesection point of p and kBC.

  4. Draw the circle k(C,P).

  5. Let D and D' be the intersections of k(C,P) and the line k.

  6. The circle through A,B,D (and A,B,D') is the solution.


Animation

|CF|2 = |FA|2 + |CD|2

The construction is based on Pythagoras' theorem for right triangles.
 
Steps:
  1. Let p be the bisector of AB.

  2. Let F be the intersection of lines p and k.

  3. Draw a semicircle kCF with CF as diameter.

  4. Let H be a point on kCF so that |FA|=|FH|

  5. Draw the circle k(C,H).

  6. Let D and D' be the intersection points of k and k(C,H).

  7. The circles through A,B,D and A,B,D' are the sought circles.

    Animation

|CA|·|CB| is the power of C wrt. the sought circle.
 

The construction is based on power theorem. In our case the power of the point C is the same for all circles passing through A,B and equals the square of the tangent segment from C to the circle.
Steps:
  1.  

  2. Let C be the intersection of line AB and k.

  3. Draw the circle with diameter AB. Let P be its centre.

  4. Draw the circle with PC as diameter.

  5. Let Q be one of the intersections of circles with diameters AB and PC.

  6. Draw the circle k(C,Q).

  7. Let D and D' be the two intersections of the line k and the circle k(C,Q).

  8. The circles through A,B,D and A,B,D' touch the line k.

The line FA is tangent to the circle through D,A,D'.
|CD| = |CD'|

 

Steps:

  1. Let C be the intersection of line k and line AB. Let F be the intersection of line k and bisector of AB.
  2. Let p be the line through A that is perpendicular to FA.

  3. Let q be the line through C, perpendicular to k.

  4. Let G be the intersection of p and q.

  5. Draw the circle k(G,A).

  6. The circle k(G,A) intersects the line k in points D and D'.

  7. The circles through A,B,D and throught A,B,D' the touch the line k.

    Animation

The centre K of the circle through G,B,G' lays on the tangent to the circle through ABF.
The line EC is perpendicular to k.

Steps:
  1. Let C be the intersection of line k and AB. Let F be the intersection of k and bisector of AB.
  2. Let D be the intersection of p and the perpendicular to k through C.
  3. Draw the line m through D that is parallel to AB.
  4. Draw the tangent n to circle ABF through B. Note that ∠(n,AB) = ∠BFA.
  5. Let K be the intersection of m and n.
  6. Let G and G' be the intersections of the circle k(K,B) with p.
  7. The circles k(G,B) and k(G',B) are the tangent to k and pass through A nad B.