Pythagorea Angle Bisectors All Levels (16.1-16.16) Solutions/Answers

Pythagorea Level 16.1 16.2 16.3 16.4 16.5 16.6 16.7 16.8 16.9 16.10 16.11 16.12 16.13 16.14 16.15 16.16 Solution/Answers

Pythagorea is android/iOS app developed by Horis International Limited. Solutions hints and answers to pythagorea are available in this post scroll down to find solutions to all the levels.

This game is mostly focused on geometric puzzles and construction. the workspace is divided into grids to draw lines. You should know all the basic Math operations. All lines and shapes are drawn on a grid whose cells are squares. Most of the game levels can be answered using natural intuition and bye some basic laws of geometry. You have to connect points on the grid using straight lines to construct an element, you can even use intersection points to draw. some levels are very easy some are of medium difficulty and some are very hard to solve, that’s why I am providing solutions to all the problems.

If you are here for levels other than ‘Angle Bisectors’ Go to directory of all other levels at :

Pythagorea Angle Bisectors (Click required Level):

Video link to these solutions (in case you want to know the detailed solutions):

Feel free to comment below if you have any doubts regarding the solutions. I will try to help you guys. All other levels are posted on this blog please visit them too.

9 thoughts on “Pythagorea Angle Bisectors All Levels (16.1-16.16) Solutions/Answers

  • September 8, 2018 at 14:55

    it is not maths, it is luck with no explaination at all.
    maths is not finding a number but understanding a process.

    • April 7, 2021 at 17:18

      Déjà 1 an et pas de réponse.
      J’ai la même demande svp, tant pis si c’est trop ?

    • April 7, 2021 at 17:20

      C’est obligatoirement des maths. C’est construit

  • June 1, 2018 at 15:22

    Can you explain the solution for Pythagorea Level 16.16? Why does it work that way?

    • January 28, 2019 at 21:38

      I struggled with 16.16 for a while and finally gave up. I’m trying to understand the logic behind the solution. I see that the line constructed in the solution, from the center of the circle, is parallel to the given ray, and therefore cuts the arc between the ray and the diameter line, perfectly in half, and therefore, the point where this line intersects the arc, gives you the bisector. But I can’t figure out how you know that to construct that parallel line, you intersect your radius line with the point defined by the intersection of that lower right hand “slice” line with the grid line.

    • February 14, 2019 at 02:25

      Don’t even need to construct anything in the “lower” part of the circle.

      Lets call the centre O at (0,0)
      Lets call the angle-point A at (sqrt(5),0)
      Lets call the “upper” given intersection B at (-1,2)
      Lets call the other given intersection C at (-sqrt(5),0)
      FYI: the angle BAC is exactly half the angle BOC
      Create point D halfway between B and C.
      OD cuts the angle COB in half.
      Lets call the new intersection between the circle and OD the point E.
      (This means the angle BOD=BOE equals BAC)
      That means OD cuts the segment CB in half at the point E.
      The angle BOE is twice the angle BAE

    • April 3, 2020 at 17:28

      If you halve the length of an arc, the angle sitting on that arc will be halved too. Now construct a rectangle with upper side of the angle as its side and with the diameter of the circle (lower side of the angle) as its diagonal. If you draw a line that splits the rectangle in half (you can see that line is drawn on the pic), it will split the arc in half too.

    • August 30, 2020 at 23:58

      The first point is 4 right and 3 up, therefore a 3-4-5 right triangle. Connect that point to the point 5 right of the vertex to make an isosceles triangle. Find the midpoint of that line segment to find the bisector.

      • August 31, 2020 at 00:01

        Wrong sorry. I was looking at 16.6 not .16


Leave a Reply

Your email address will not be published.