    # Learn Geometry with GeoGebra

GeoGebra is a free dynamic mathematics software and it is especially created for teaching and learning. It is an open-source software available for multiple platforms (iOS, Android, Windows, Mac OS, and Linux) and it has been translated into many languages. It has an easy-to-use interface and yet many powerful features.

GeoGebra is both an online web application and an online downloadable application.

## Drawing lines and segments

You only need to use the appropriate tools in the Toolbar to create geometric constructions on the Graphics View with your mouse. For example, select Line, click twice on the Graphics View and you will get two points (A, B) and a line.

If you want to draw segments, select the Segment option on the Line menu. Click again twice on the Graphics View and you will get a segment. It is also possible to get the same results from the input field in the bottom panel. If you want to introduce points, just type: A= (1, 6). B = (3, 2). Drawing lines is very simple, too. Let’s draw the line that passes through two points. Let’s find its slope = (ya-yb)/(xa-xb) = (6-2)/(1-3)=-2. Now we can write its equation: y-ya = slope*(x-xa), (y-6)=-2*(x-1), (y-6)=-2x+2, y=-2x+2+6, so you need to type, y=-2x+8.

Draw a line (A, B) and a point (C). Let’s construct _the parallel line to our initial line which passes through C. Select the Parallel Line option on the Perpendicular Line menu. Click on C, and then on our previous line (A, B). Let’s construct the perpendicular line to our initial line which passes through C. Just select Perpendicular Line, click on C, and then on our previous line (A, B). The algebra view on the left panel allows you to see and edit all created objects. For instance, point B and its two coordinates.

Next, we will measure the length of a segment. Draw two points (A, B) and a segment (AB). Select Angle, Distance or Length, and click on the segment. You can also type commands in the Input field and get the same results. For example, d = Distance [A, B] calculates the length of the segment AB; PerpendicularLine[C, a] creates a line through the point C perpendicular to the given line (a); and Line[C, a] creates a line through the point C parallel to the given line (a).

Let’s calculate the centre of our segment (AB). Select the Midpoint or Center option on the Point menu, and click on the segment. Alternatively, you can introduce the command Midpoint [A, B].

## Drawing angles, circles, tangents, and bisectors.

Let’s draw an angle and its bisector.

• Draw three points (A, B, and C), select Angle in the Toolbox and finally, click on C, A, and B. The order is important because Geogebra works in a counter clockwise orientation. You can also do this by typing the command: Angle (C, A, B).
• How to construct a bisector (it divides an angle into two congruent angles) of a given angle? Select the Angle Bisector option on the Perpendicular Line menu, then click on C, A, and B. It is very easy, isn’t it?

Geogebra is a dynamic software application, so you can move all the points A, B, or C, and the bisector will update automatically. Let’s draw a circle.

Select Circle with Center Through Point in the Toolbox and then, click on one point in the centre panel (centre, A) and then another (one point in the circle, B). As always, you can construct circles by typing myCircle = Circle [A, B] in the input field.

Given a circle and a point C, we will draw the two tangent lines to our circle which pass through C. Select Perpendicular Line, Tangents, and click on point C and the circle or just type, Tangent [myCircle, C]. # Let’s construct the midpoint of a segment

• Draw a segment “a”, select the Midpoint or Center option from the New Point menu, and click on segment “a” or just type C = Midpoint[A, B] in the Input text box. Let’s check that it is correct! The midpoint (x, y) of a segment with endpoints (x(A), y(A)) and (x(B), y(B)) has coordinates: ((x(A)+ x(B))/2, (y(A)+ y(B))/2) = ( (2.36 + 11.06)/2, (-2.67 + -0.97)/2 )= (6.71, -1.82).
• Draw two circles with centers A and B respectively and the same radius. This radius should be greater than the length of the segment AC. In other words, r = Distance[A, C] = 4.43, c = Circle[A, r’], d = Circle[B, r’], with r’ > r (eg. 6 > 4.43).
• Go to New Point, Intersect two objects and select both circles (c, d) or type Intersect[c,d], Geogebra will return two points D and E. Then, draw the line which passes through these two points: Segment[D, E] or select Line Through two Points, and click on D and E. Observe that this line passes through our Midpoint C, and by doing so, it confirms that C is the midpoint of our original segment AB.

# Calculate the area of a circle

• Select Circle with Center Through Point in the Toolbox and then, click on one point in the center panel (center, A) and then another (one point in the circle, B).
• Go to Angle, Area and click on the circle. This tool gives you the area of a polygon, circle, or ellipse, and the alternative command is myArea = Area[c] where c is the circle.
• Let’s draw the radius: Line through Two Points, Segment between Two Points, and click on A and B. Next, select from the Angle menu, the Distance or Length option and click on A and B.
• Finally, let’s calculate the area algebraically: Pi*r2 = Pi*2.962 = 27.53. There are approximation errors involved. Let’s calculate it with Python:
``````user@pc:~\$python
Python 3.8.6 (default, May 27 2021, 13:28:02)
[GCC 10.2.0] on linux
import math
math.pi * pow(2.96, 2)
27.52537819369233
`````` # Working with Triangles

• First, let’s view the grid (View, Grid) and maybe increase the font size (Options, Font Size, 20pt) and Zoom In (Move Graphics View).
• Let’s draw a triangle. Go to the Polygon menu and select three vertices: A = (1, 1), B = (1, 4), and C = (5, 1), and click back again on the point A to close the polygon.
• Next, let’s draw its three angles and demonstrate that the sum of the measures of the interior angles of a triangle is 180°. Select Angle and click on C, A, B (the order is important, you always need to click on the points in a counterclockwise direction): α = 90°. Repeat with B, C, A (β = 36.87°) and A, B, C (γ = 53.13°). Observe that α + β + γ = 90° + 36.87° + 53.13° = 180. Type in the input text area: ζ = α + β + γ and select Insert Text: α + β + γ (Symbols) = α + β + γ = ζ (Objects).
• The Pythagorean theorem states that the square of the hypotenuse side is equal to the sum of squares of the other two sides. Select the Distance or Length option from the Angle menu and click on B and C: |BC| = 5. Obviously, |BC|2 =|BA|2 + |CA|2 =32 + 42 = 25.
• Calculate the area by clicking on Angle, Area and the triangle. Area ABC = 6. Remember that the area of a triangle is equal to half of the base times height = (|BA| * |CA|)⁄2 = (3 * 4)⁄2 = 6. # The Circumscribed circle

The circumscribed circle of a polygon is a circle that passes through all the vertices of the polygon. Its center is called the circumcenter. Not every polygon has a circumscribed circle. A polygon that does have one is called a cyclic polygon. All triangles are cyclic.

• The interior perpendicular bisector of a side of a triangle is the segment that perpendicularly bisects that side. The three perpendicular bisectors of a triangle’s three sides intersect at the circumcenter (d, e, f). You can draw them by selecting Perpendicular Line, Perpendicular Bisector and clicking on each of the triangle’s sides or typing PerpendicularBisector[a], PerpendicularBisector[b], and PerpendicularBisector[c] where a, b, and c are the triangle’s sides.
• Draw a circle by selecting Circle with Centre through Point and clicking on the circumcenter and on any of the three triangle’s vertices or just typing the command Circle[ Intersect[d, e], A] # The incircle or inscribed circle

The incircle or inscribed circle of a triangle is the largest circle contained in the triangle. It touches or is tangent to the three sides. The center of the incircle is called the triangle’s incenter.

• The center of the triangle’s incircle (the incenter) can be found as the intersection of the three internal angle bisectors. Select in the menu Perpendicular Line, the option Angle Bisector, and click on C, A, and B (d). Repeat with B, C, and A (e), and do it again with C, B, and A (f) or type d = AngleBisector[C, A, B], e = AngleBisector[B, C, A], and f = AngleBisector[C, B, A] in the Input text area.
• Type Incenter = Intersect[d, e] (or Incenter = Intersect[e, f] or Incenter = Intersect[d, f]) or select New Point, Intersect Two Objects and click on two angle bisectors.
• Write D = Intersect[d, Segment[B, C ]] where d is the angle bisector of C, A, and B or select New Point, Intersect Two Objects and click on d and the segment BC. Alternatively, you could use the intersection between the Angle Bisector of C, B, A (f) and the side AC or between the Angle Bisector of B, C, A (e) and AB.
• Draw the incircle by selecting Circle with Centre through Point and clicking on the triangle’s incircle (Incenter) and D (the intersection between the Angle Bisector of C, A, and B (d) and the side BC). # Translations

There are three basic transformations: translations, reflections, and rotations. These transformations move the figure or object without making any changes to its shape and size. A reflection flips an object over to create a mirror image, a rotation turns an object, and a translation moves or slides an object to a different location without rotating or resizing it. I want to insist again that after any of these transformations (flip, turn or slide) the figure or object still has the same size, area, angles, and line lengths.

• Let’s draw a point (New Point, A) and a vector (Line through Two Points, Vector between Two Points, u = Vector [B, C]).
• Next, select the option Translate Object by Vector from the menu Reflect Object in Line, and click on our point (A) and the vector (u). The translated point is A’ = (7.66, 0.24) = A + u = (1.74, -2.46) + (5.92, 2.7).
• Similarly, we can translate triangles: Translate[poly1, u] where poly1 is the triangle. Observe: D’ = Translate[D, u] = D + u = (4.56, -4.24) + (5.92, 2.7) = (10.48, -1.54).

Calculate the area (Angle, Area) and the triangle’s interior angles. When you translate something, you’re simply moving it around. You don’t distort it in any way. If you translate a segment, its length doesn’t change. Similarly, if you translate an angle or a triangle, the measure of the angle or the triangle’s area do not change. # Rotations in Geogebra

• Let’s draw a square. Select the option Regular Polygon from the Polygon menu, click on two points A and B and specify 4 as the number n of vertices in the input field of the appearing dialog window.

• Our center point of rotation (New Point, O) is going to be (0, 0), the origin where the axes of the Cartesian system intersect.

• Next, we will rotate the square 45° degrees around the origin. Go to Reflect Object in Line, Rotate Object around Point by Angle and click on our square (poly1), the centre point (O) and enter the angle 45°.

• Check your results. Observe the rotation matrix, the transformation matrix that is used to perform a rotation. We are also using Overleaf, an online LaTeX editor. A (x, y), A’ (x’, y’) is the point A rotated around O 45°, x’ = 1.9 * cos(45 degrees -not radians-) -3.36 * sin(45 degrees)=-1.03.  Calculate the area (Angle, Area) and the square’s interior angles. When you rotate something, you’re simply whirling it around. You don’t distort it in any way and its area does not change.

# Reflection

A reflection can be thought of as flipping an object. An object and its reflection do have the same shape and size, but they appear as if they are mirror reflections. Every point is the same distance from the central or the mirror line.

• Draw a triangle (Polygon, poly1 = Polygon[A, B, C]), and a Line through Two Points (-2, 0) and (4, 0); in other words, the x axis (d: y = 0).
• Let’s reflect the triangle across the x-axis. Select Reflect Object in Line, then click on poly1 and d (d: y = 0, the x axis).
• Take into account that when you reflect a point A = (x, y) across the x-axis, the x-coordinate remains the same, unaltered but the y-coordinate is negated (transformed into its opposite, its sign is changed) A’ = (x, -y) = (2.58, -4.18). Next, we will reflect the triangle across the y-axis, so the line d is x = 0. Observe that the y-coordinate remains the same, unaltered but the x-coordinate is negated (transformed into its opposite) A’ = (-x, y) = (-2.58, 4.18). Let’s reflect the triangle across the line y = x. When you do so, the x-coordinate and y-coordinate swap or change places A’ = (y, x) = (4.18, 2.58). Finally, draw a pentagon (Polygon, Regular Polygon, poly1), a point (New Point, F(0,0)), and reflect the pentagon over the origin by selecting from the menu Reflect Object in Line, the option Reflect Object in Point, and clicking on the pentagon and the point F. Observe that both the x-coordinate and the y-coordinate are negated (transformed into their opposite), so A’ = (-x, -y) = (-2.73, -3.28) Bitcoin donation 