On Using GeoGebra and ChatGPT for Geometric Discovery

Appendix A.1.1. Descriptive Statement Prompt

 Yes, we can prove this statement using vector geometry. Let’s consider a quadrilateral $ABCD$ and denote the midpoints of $AB$, $CD$, $AC$, and $BD$ as E, F, G, and H respectively.

 1. Midpoints of Opposite Sides:

•  Let E be the midpoint of $AB$.
•  Let F be the midpoint of $CD$.

 2. Midpoints of Diagonals:

•  Let G be the midpoint of $AC$.
•  Let H be the midpoint of $BD$.

 We need to prove that $EFGH$ forms a parallelogram.

 Step-by-Step Proof:

 Step 1: Define the Points Using Vectors

•  Let the position vectors of points A, B, C, and D be $\overrightarrow{A}$, $\overrightarrow{B}$, $\overrightarrow{C}$, and $\overrightarrow{D}$ respectively.

 Step 2: Find Midpoints Using Vectors

•  The position vector of E, the midpoint of $AB$, is given by:

  $$\overrightarrow{E}={\displaystyle \frac{\overrightarrow{A}+\overrightarrow{B}}{2}}$$
•  The position vector of F, the midpoint of $CD$, is given by:

  $$\overrightarrow{F}={\displaystyle \frac{\overrightarrow{C}+\overrightarrow{D}}{2}}$$
•  The position vector of G, the midpoint of $AC$, is given by:

  $$\overrightarrow{G}={\displaystyle \frac{\overrightarrow{A}+\overrightarrow{C}}{2}}$$
•  The position vector of H, the midpoint of $BD$, is given by:

  $$\overrightarrow{H}={\displaystyle \frac{\overrightarrow{B}+\overrightarrow{D}}{2}}$$

 Step 3: Calculate Vectors of the Sides of Quadrilateral

•  Vector $EF$:

  $$\overrightarrow{EF}=\overrightarrow{F}-\overrightarrow{E}={\displaystyle \frac{\overrightarrow{C}+\overrightarrow{D}}{2}}-{\displaystyle \frac{\overrightarrow{A}+\overrightarrow{B}}{2}}={\displaystyle \frac{\overrightarrow{C}+\overrightarrow{D}-\overrightarrow{A}-\overrightarrow{B}}{2}}$$
•  Vector $GH$:

  $$\overrightarrow{GH}=\overrightarrow{H}-\overrightarrow{G}={\displaystyle \frac{\overrightarrow{B}+\overrightarrow{D}}{2}}-{\displaystyle \frac{\overrightarrow{A}+\overrightarrow{C}}{2}}={\displaystyle \frac{\overrightarrow{B}+\overrightarrow{D}-\overrightarrow{A}-\overrightarrow{C}}{2}}$$

 Notice that:

 Since $\overrightarrow{EF}$ and $\overrightarrow{GH}$ are equal, $EF$ is parallel and equal in length to $GH$.

• Vector $FG$:

  $$\overrightarrow{FG}=\overrightarrow{G}-\overrightarrow{F}={\displaystyle \frac{\overrightarrow{A}+\overrightarrow{C}}{2}}-{\displaystyle \frac{\overrightarrow{C}+\overrightarrow{D}}{2}}={\displaystyle \frac{\overrightarrow{A}-\overrightarrow{D}}{2}}$$
• Vector $EH$:

  $$\overrightarrow{EH}=\overrightarrow{H}-\overrightarrow{E}={\displaystyle \frac{\overrightarrow{B}+\overrightarrow{D}}{2}}-{\displaystyle \frac{\overrightarrow{A}+\overrightarrow{B}}{2}}={\displaystyle \frac{\overrightarrow{D}-\overrightarrow{A}}{2}}$$

 Notice that:

 Since $\overrightarrow{FG}$ and $\overrightarrow{EH}$ are equal, $FG$ is parallel and equal in length to $EH$.

 Conclusion: Both pairs of opposite sides $EF$ and $GH$, $FG$ and $EH$ are equal and parallel. Therefore, the quadrilateral $EFGH$ is a parallelogram.

 This completes the proof.

Appendix A.1.2. Figural Prompt

Figure A1. Image describing context of Theorem 1.

Figure A1. Image describing context of Theorem 1.

 To prove that EFGH is a parallelogram in the given figure, we’ll use the properties of midpoints and vectors as previously outlined. Let’s identify and label the points clearly from the image:

•  A, B, C, and D are the vertices of the quadrilateral.
•  E is the midpoint of $AB$.
•  G is the midpoint of $CD$.
•  F is the midpoint of $BC$.
•  H is the midpoint of $AD$.

 Step-by-Step Proof:

 Step 1: Define the Points Using Vectors

 Let’s assign position vectors to points A, B, C, and D as $\overrightarrow{A}$, $\overrightarrow{B}$, $\overrightarrow{C}$, and $\overrightarrow{D}$ respectively.

 Step 2: Find Midpoints Using Vectors

•  The position vector of E, the midpoint of $AB$, is given by:

  $$\overrightarrow{E}={\displaystyle \frac{\overrightarrow{A}+\overrightarrow{B}}{2}}$$
•  The position vector of G, the midpoint of $CD$, is given by:

  $$\overrightarrow{G}={\displaystyle \frac{\overrightarrow{C}+\overrightarrow{D}}{2}}$$
•  The position vector of F, the midpoint of $BC$, is given by:

  $$\overrightarrow{F}={\displaystyle \frac{\overrightarrow{B}+\overrightarrow{C}}{2}}$$
•  The position vector of H, the midpoint of $AD$, is given by:

  $$\overrightarrow{H}={\displaystyle \frac{\overrightarrow{A}+\overrightarrow{D}}{2}}$$

 Step 3: Calculate Vectors of the Sides of Quadrilateral $EFGH$

•  Vector $EF$:

  $$\overrightarrow{EF}=\overrightarrow{F}-\overrightarrow{E}={\displaystyle \frac{\overrightarrow{B}+\overrightarrow{C}}{2}}-{\displaystyle \frac{\overrightarrow{A}+\overrightarrow{B}}{2}}={\displaystyle \frac{\overrightarrow{C}-\overrightarrow{A}}{2}}$$
•  Vector $GH$:

  $$\overrightarrow{GH}=\overrightarrow{H}-\overrightarrow{G}={\displaystyle \frac{\overrightarrow{A}+\overrightarrow{D}}{2}}-{\displaystyle \frac{\overrightarrow{C}+\overrightarrow{D}}{2}}={\displaystyle \frac{\overrightarrow{A}-\overrightarrow{C}}{2}}$$

 Notice that:

 Since $\overrightarrow{EF}$ and $\overrightarrow{GH}$ are equal in magnitude and direction, $EF$ is parallel to $GH$ and equal in length to $GH$.

•  Vector $FG$:

  $$\overrightarrow{FG}=\overrightarrow{G}-\overrightarrow{F}={\displaystyle \frac{\overrightarrow{C}+\overrightarrow{D}}{2}}-{\displaystyle \frac{\overrightarrow{B}+\overrightarrow{C}}{2}}={\displaystyle \frac{\overrightarrow{D}-\overrightarrow{B}}{2}}$$
•  Vector $EH$:

  $$\overrightarrow{EH}=\overrightarrow{H}-\overrightarrow{E}={\displaystyle \frac{\overrightarrow{A}+\overrightarrow{D}}{2}}-{\displaystyle \frac{\overrightarrow{A}+\overrightarrow{B}}{2}}={\displaystyle \frac{\overrightarrow{D}-\overrightarrow{B}}{2}}$$

 Notice that:

 Since $\overrightarrow{FG}$ and $\overrightarrow{EH}$ are equal in magnitude and direction, $FG$ is parallel to $EH$ and equal in length to $EH$.

 Conclusion: Both pairs of opposite sides $EF$ and $GH$, $FG$ and $EH$ are equal in length and parallel. Therefore, quadrilateral $EFGH$ is a parallelogram.

Appendix A.2.1. Descriptive Statement Prompt

 To prove the given statement about a trapezoid, we’ll consider a trapezoid $ABCD$ with $AB$ and $CD$ as the parallel sides ($AB\Vert CD$). Let the non-parallel sides be $AD$ and $BC$.

 We need to prove that the line passing through the midpoints of the diagonals $AC$ and $BD$ intersects the non-parallel sides $AD$ and $BC$ at their midpoints.

 Here is the proof:

 1. Identify the Midpoints of the Diagonals: Let E be the midpoint of diagonal $AC$ and F be the midpoint of diagonal $BD$.

 2. Coordinate Geometry Setup: Place the trapezoid $ABCD$ in a coordinate system with $A=(0,0)$, $B=(b,0)$, $C=(c,d)$, and $D=(a,d)$.

 3. Find Coordinates of the Midpoints E and F:

 The coordinates of midpoint E of diagonal $AC$ are:

 The coordinates of midpoint F of diagonal $BD$ are:

 4. Equation of Line through E and F:

 The slope m of the line passing through points E and F is:

 Therefore, the line through E and F is horizontal, given by $y={\displaystyle \frac{d}{2}}$.

 5. Midpoints of Non-Parallel Sides:

 The midpoint G of $AD$ (where $A=(0,0)$ and $D=(a,d)$) is:

 The midpoint H of $BC$ (where $B=(b,0)$ and $C=(c,d)$) is:

 6. Verify Intersection Points:

 Since both midpoints G and H lie on the horizontal line $y={\displaystyle \frac{d}{2}}$, which is the same as the line through E and F, the line through the midpoints of the diagonals intersects the non-parallel sides $AD$ and $BC$ at their midpoints G and H.

 Therefore, we have proven that the line passing through the midpoints of the diagonals $AC$ and $BD$ intersects the non-parallel sides $AD$ and $BC$ at their respective midpoints.

Appendix A.2.2. Figural Prompt

 To prove that segment $MB$ is equal to segment $MA$ in the given trapezoid $ABCD$ with $AB\Vert CD$, let’s use a combination of coordinate geometry and properties of midpoints in trapezoids. The key insight is to show that M is the midpoint of $AB$.

 Step-by-Step Proof:

 1. Identify the Key Points:

•  Let A be at $(0,0)$, B at $(b,0)$, C at $(c,d$), and D at $(a,d)$.
•  Let E be the midpoint of diagonal $AC$ and F be the midpoint of diagonal $BD$.
•  Let M and K be the midpoints of $AB$ and $CD$, respectively.

 2. Find Coordinates of Midpoints M and K:

•  M, the midpoint of $AB$:

  $$M=\left({\displaystyle \frac{0+b}{2}},{\displaystyle \frac{0+0}{2}}\right)=\left({\displaystyle \frac{b}{2}},0\right)$$
•  K, the midpoint of $CD$:

  $$K=\left({\displaystyle \frac{c+a}{2}},{\displaystyle \frac{d+d}{2}}\right)=\left({\displaystyle \frac{c+a}{2}},d\right)$$

 3. Equation of Line through M and K:

•  The slope m of the line $MK$ is:

  $$m={\displaystyle \frac{d-0}{\frac{c+a}{2}-\frac{b}{2}}}={\displaystyle \frac{d}{\frac{c+a-b}{2}}}={\displaystyle \frac{2d}{c+a-b}}$$
•  The equation of line $MK$ can be written using point $M=({\displaystyle \frac{b}{2}},0)$:

  $$y={\displaystyle \frac{2d}{c+a-b}}\left(x-{\displaystyle \frac{b}{2}}\right)$$

 4. Prove $MB=MA$:

•  Since M is the midpoint of $AB$, we have: $MB=MA$ because midpoints divide a line segment into two equal parts by definition.

 This implies that, geometrically, if M is the midpoint of $AB$, then segments $MB$ and $MA$ are congruent as they are each half of $AB$.

 Conclusion: In the given trapezoid $ABCD$, with $AB\Vert CD$ and M being the midpoint of $AB$, we have $MB=MA$. This is because M divides $AB$ into two equal segments by definition, confirming that $MB=MA$.