### How is an object related to a subset of objects in a construction? How to construct an object from given objects?

Given a configuration, OK Geometry lists the observed properties. With subsequent query buttons (magenta coloured buttons on the right hand side of the displayed construction, e.g. ?⚫, ?△ ) you ask OK Geometry to display only properties that contain a particular object.

In Plus mode, the query buttons have additional functionality: in a configuration they attempt to relate a specified object to a specified subset of objects (called given objects) in a configuration, and also with many objects geometrically derived from the specified subset of objects.

The advanced query is a great help for hypothetising solutions to construction tasks. It also reveals surprising new relations among objects in a configuration.

A detailed description of the use of advanced query can be found here.

We illustrate the power of advanced query with some examples.

### Example 1

 Inscribe a square PQRS into a given triangle ABC as shown in the figure. In the configuration, which can be obtained by circumscribing a triangle ABC to a square PQRS, we declare (green ⚫? button) as non-given the points P,Q,R,S as well as all sides of the square. Then we apply Observe command and Advanced query command (magenta ?⚫ button) on non-given points, e.g. R and Q. On the bottom of the lists of properties that include the queried point we get several indication for the solution. Below we present only some of them.  The point R lays on the line through the points Y353 and Y360 (or _353, _360; note that the indexes vary), where Y353 is the image of A rotated by 90 degrees clockwise around B, Y360 is the orthogonal projection of C onto AB. The point Q lays on the line through C and Y280 where Y280 is the image of A rotated counterclockwise by 90 degrees around B.

### Example 2

In the following example we investigate how to construct a circle that is tangent to three specific objects. In such cases it is often a good idea to determine first the points of tangency of the circle to be constructed.

 We have a circle sector with centre S and AB as arc and an additional arc AC, which has the midpoint T of AS as centre, while the endpoint C of the arc is on BS. We want to inscribe a circle into the triangular patch between the line BS and the arcs AC and AB. The inscribed circle is easily drawn with OK Geometry, using the command Circle 3 objects. To get hints for a construction we mark the inscribed circle, the touch points D,E,F, and the centre I of the inscribed circle as non-given objects. Then we proceed with the command Observe and, finally, we query the points. Here is one of the several suggestions for obtaining the point D: The point D lays on the line through the auxiliary points Y1486 and Y819, where Y1486 is the intersection of the circle k(S,A) and the bisector of AB; Y819 ia the image of A rotated counterclocwise by 90 degrees around B. Note that the described line does not contain the point C (as it may appear in the figure). Once the point D is known, it is a good idea to mark it as known, and repeat the Observe procedure. The advanced query of the centre I provides several suggestions for the construction of I (obviously I lays on the segment SD). Here is one of them: The point lays on the line through Y4011 and Y1897, where Y4011 is the syimmedian point of the triangle ADS, and Y1897 is the mirror image in point D of the midpoint of AD. ### Example 3

The following example is the 25th problem of the famous Wernick list. Construct a triangle ABC if the location of the following points of the triangle ABC are known:
• the vertex A,
• the midpoint Ma of the side BC,
• the intersection Ta of the angle bisector at A and the side BC.

 In a triangle ABC we construct the points Ma and Ta. Note that A, Ma, Ta, and the line BC are given (as well as the location of the centroid). We mark everything else as non-given. Since observation command and query of points B, C yields nothing useful, we try to determine some other important point of the triangle.The non-given circumcentre O is a good choice. An observation and query of the circumcentre O yields several hypotheses, which can be quite easyly proved and used to solve the construction problem. Here is the first (trivial property): the O(Ma) is perpendicular to BC at Ma. Furthermore, O lays on the bisector of the segment A(Y364), where Y364 is the antipode of Ta in the circle through A,Ta,Ma. The circumcentre O also lays on the circle through points A, Ma, Y17, where Y17  is the intersection of line BC with the bisector of A(Ta). Note that O is the antipode of Y17 in the mentioned circle, and the circle contains also points Y21 - the intersection of A(Ta) with the bisector of (Ma)(Ta), Y6 - the inersection of BC with the bisector of (Ma)(Ta). 