9- Geometrical constructions

# 9- Geometrical constructions ## Constructing the perpendicular bisector of a line ### Introduction Welcome to this lesson on geometrical constructions! Today, we will learn how to accurately construct the perpendicular bisector of any given line segment. This is a fundamental skill in geometry that is used in many other constructions and in solving various geometric problems. Understanding how to create a line that is both perpendicular to another line segment and divides it into two equal halves is crucial for your mathematical journey. ### Key Concepts #### What is a Perpendicular Bisector? A perpendicular bisector of a line segment is a line that meets two conditions: 1. **Perpendicular:** It intersects the line segment at a \( 90^ \) degree angle. 2. **Bisector:** It divides the line segment into two equal parts, meaning it passes through the midpoint of the segment. #### Properties of a Perpendicular Bisector * It always passes through the midpoint of the line segment. * It forms a right angle (\( 90^ \)) with the line segment. * Any point on the perpendicular bisector is equidistant (the same distance) from the two endpoints of the line segment. This property is key to understanding why our construction method works. #### Tools Required To construct a perpendicular bisector, you will need: * A sharp pencil * A ruler (or straightedge) * A compass #### Steps for Constructing a Perpendicular Bisector Follow these steps carefully to construct the perpendicular bisector of a given line segment \( \overline{AB} \): 1. **Draw the Line Segment:** Using your ruler, draw the line segment \( \overline{AB} \) that you want to bisect. 2. **Set Compass Radius:** Place the compass point on one endpoint, say point A. Open the compass to a radius that is **more than half** the length of \( \overline{AB} \). This is important to ensure your arcs intersect. 3. **Draw Arcs from First Endpoint:** With the compass point on A and the radius set, draw an arc above and an arc below the line segment \( \overline{AB} \). 4. **Draw Arcs from Second Endpoint:** Without changing the compass radius, move the compass point to the other endpoint, point B. Draw another arc above and an arc below the line segment, ensuring these arcs intersect with the arcs you drew from point A. 5. **Connect Intersection Points:** You should now have two points where the arcs intersect (one above \( \overline{AB} \) and one below \( \overline{AB} \)). Label these intersection points, say C and D. Use your ruler to draw a straight line connecting point C and point D. This line \( \overline{CD} \) is the perpendicular bisector of \( \overline{AB} \). #### Justification of the Construction This construction works because of the property that any point on the perpendicular bisector is equidistant from the endpoints of the line segment. When you draw the arcs from A and B with the same radius: * Any point on the arc drawn from A is equidistant from A (with a distance equal to the compass radius). * Any point on the arc drawn from B is equidistant from B (with the same compass radius). * Therefore, the two intersection points (C and D) are both equidistant from A and B. Since C is equidistant from A and B (\( AC = BC \)), and D is also equidistant from A and B (\( AD = BD \)), the line \( \overline{CD} \) must contain all points equidistant from A and B. This line is precisely the perpendicular bisector of \( \overline{AB} \). ### Vocabulary * **Perpendicular:** Forming a right angle (\( 90^ \)) with another line or surface. * **Bisect:** To divide something into two equal parts. * **Line Segment:** A part of a line that has two distinct endpoints. * **Compass:** A geometric tool used for drawing circles or arcs and for measuring distances. * **Straightedge:** A tool with a straight edge, used to draw straight lines (a ruler without markings). * **Arc:** A part of the circumference of a circle or any curved line segment. * **Midpoint:** The point that divides a line segment into two equal parts. * **Equidistant:** At equal distances. ### Examples #### Example 1: Constructing the Perpendicular Bisector of a Line Segment 7 cm Long **Problem:** Construct the perpendicular bisector of a line segment \( \overline{PQ} \) where \( PQ = 7 \text{ cm} \). **Solution Steps:** 1. Draw a line segment \( \overline{PQ} \) that is 7 cm long using your ruler. 2. Place the compass point on P. Open the compass to a radius greater than half of 7 cm (e.g., 4.5 cm). 3. Draw an arc above and an arc below \( \overline{PQ} \). 4. Without changing the compass opening, place the compass point on Q. Draw another arc above and an arc below \( \overline{PQ} \) such that they intersect the first set of arcs. 5. Label the intersection points R (above) and S (below). Use your ruler to draw a straight line connecting R and S. This line \( \overline{RS} \) is the perpendicular bisector of \( \overline{PQ} \). 6. You can verify by measuring: the line \( \overline{RS} \) should cut \( \overline{PQ} \) exactly at its midpoint (3.5 cm from P and Q), and if you use a protractor, the angle of intersection should be \( 90^ \). ### Exercises 1. Construct the perpendicular bisector of a line segment \( \overline{AB} \) that is 6 cm long. 2. Draw a line segment \( \overline{XY} \) of length 9 cm. Construct its perpendicular bisector. 3. Explain in your own words why the compass radius must be more than half the length of the line segment during construction. 4. If a point M lies on the perpendicular bisector of line segment \( \overline{CD} \), what can you say about the lengths of \( \overline{MC} \) and \( \overline{MD} \)? ### Assignment 1. Draw an acute-angled triangle \( \triangle ABC \). Construct the perpendicular bisector of each side (\( \overline{AB} \), \( \overline{BC} \), and \( \overline{CA} \)). What do you observe about the intersection of these three perpendicular bisectors? 2. Construct a line segment \( \overline{EF} \) of length 8.5 cm. Construct its perpendicular bisector. Then, construct the perpendicular bisector of one of the halves (e.g., the segment from E to the midpoint). 3. A farmer wants to place a water trough equidistant from two trees, Tree A and Tree B, in his field. Using a diagram, show how the farmer can find the possible locations for the trough using a geometric construction. ### Summary In this lesson, we have learned the essential skill of constructing a perpendicular bisector of a line segment. We understood that a perpendicular bisector divides a line segment into two equal parts and forms a right angle with it. The construction method relies on the property that any point on the perpendicular bisector is equidistant from the segment's endpoints. By following the steps precisely with a compass and straightedge, you can accurately construct a perpendicular bisector, a fundamental tool in geometry. ## Constructing a triangle: Of given dimensions, ### Introduction Welcome to this lesson on geometrical constructions! In mathematics, particularly geometry, we often need to draw shapes accurately using only a few basic tools. Today, we will focus on constructing triangles when specific dimensions (sides and angles) are provided. Mastering these construction techniques is fundamental for understanding more complex geometrical problems and for practical applications in fields like architecture and engineering. ### Key Concepts Geometrical constructions involve drawing geometric figures using only a compass and a straightedge (ruler without markings). A protractor can also be used to measure and draw angles accurately. #### Tools for Construction * **Compass:** Used to draw circles or arcs and to transfer lengths. * **Straightedge (Ruler):** Used to draw straight lines. * **Protractor:** Used to measure and draw angles. #### Case 1: Constructing a Triangle Given Three Sides (SSS) This method is used when the lengths of all three sides of the triangle are known. For a triangle with sides \(a\), \(b\), and \(c\), the triangle inequality theorem states that the sum of the lengths of any two sides must be greater than the length of the third side. That is: \[ a + b > c \] \[ a + c > b \] \[ b + c > a \] **Steps:** 1. Draw the longest side of the triangle as the base using your straightedge. 2. Set your compass to the length of the second side. Place the compass point at one endpoint of the base and draw an arc above the base. 3. Set your compass to the length of the third side. Place the compass point at the other endpoint of the base and draw another arc to intersect the first arc. 4. The point where the two arcs intersect is the third vertex of the triangle. Connect this point to the endpoints of the base using your straightedge. #### Case 2: Constructing a Triangle Given Two Sides and the Included Angle (SAS) This method is used when the lengths of two sides and the measure of the angle between them (the included angle) are known. **Steps:** 1. Draw one of the given sides as the base using your straightedge. 2. At one endpoint of the base, use your protractor to draw the included angle. Draw a line segment of appropriate length along this angle. 3. Measure the length of the second given side along the line you just drew in step 2. Mark this point. 4. Connect the endpoint of the base (where the angle was drawn) to the marked point from step 3. This completes the triangle. #### Case 3: Constructing a Triangle Given Two Angles and One Side (ASA) This method is used when the measures of two angles and the length of the side between them (the included side) are known. Remember that the sum of angles in a triangle is always \(180^ ext{o}\). **Steps:** 1. Draw the given side as the base using your straightedge. 2. At one endpoint of the base, use your protractor to draw the first given angle. Draw a line extending from this point. 3. At the other endpoint of the base, use your protractor to draw the second given angle. Draw a line extending from this point. 4. The point where the two lines drawn in steps 2 and 3 intersect is the third vertex of the triangle. The triangle is now complete. #### Case 4: Constructing a Right-Angled Triangle Given the Hypotenuse and One Other Side (RHS) This method is a special case for right-angled triangles where the hypotenuse (the side opposite the right angle) and one other side are known. We will construct a \(90^ ext{o}\) angle. **Steps:** 1. Draw the given non-hypotenuse side as the base using your straightedge. 2. At one endpoint of the base, construct a perpendicular line (a \(90^ ext{o}\) angle). You can do this with a protractor or using a compass and straightedge (by drawing arcs to find a perpendicular bisector, or constructing a right angle at a point on a line). 3. Set your compass to the length of the hypotenuse. Place the compass point at the *other* endpoint of the base (where the perpendicular was *not* drawn). 4. Draw an arc that intersects the perpendicular line you drew in step 2. The point of intersection is the third vertex of the triangle. 5. Connect the endpoint from step 3 to the intersection point to complete the hypotenuse and the triangle. ### Vocabulary * **Triangle:** A polygon with three sides and three angles. * **Side:** One of the line segments forming the boundary of a triangle. * **Angle:** The space (measured in degrees or radians) between two intersecting lines or surfaces at or near the point where they meet. * **Compass:** A geometric tool used for drawing circles or arcs and measuring distances. * **Straightedge:** A tool used for drawing straight lines; typically a ruler without markings for measurement. * **Protractor:** A tool used for measuring and drawing angles. * **Included Angle:** The angle between two given sides of a triangle. * **Included Side:** The side between two given angles of a triangle. * **Hypotenuse:** The longest side of a right-angled triangle, opposite the right angle. * **Perpendicular:** Lines, segments, or rays that intersect to form a right angle (\(90^ ext{o}\)). * **Arc:** A part of the circumference of a circle or any curved part of a path or trajectory. * **Vertex (plural: Vertices):** A point where two or more lines, rays, or edges meet. ### Examples #### Example 1: Constructing a Triangle (SSS) Construct a triangle ABC with sides AB = 7 cm, BC = 5 cm, and AC = 6 cm. **Solution:** 1. Draw a line segment AB of length 7 cm. 2. With A as the center and radius 6 cm, draw an arc. 3. With B as the center and radius 5 cm, draw another arc to intersect the first arc at point C. 4. Join A to C and B to C. Triangle ABC is formed. #### Example 2: Constructing a Triangle (SAS) Construct a triangle PQR with PQ = 6 cm, QR = 4 cm, and angle PQR = \(60^ ext{o}\). **Solution:** 1. Draw a line segment PQ of length 6 cm. 2. At point Q, use a protractor to draw an angle of \(60^ ext{o}\). Label the line forming this angle as QX. 3. Along QX, mark a point R such that QR = 4 cm. 4. Join P to R. Triangle PQR is formed. #### Example 3: Constructing a Triangle (ASA) Construct a triangle XYZ with XY = 8 cm, angle YXZ = \(45^ ext{o}\), and angle XYZ = \(70^ ext{o}\). **Solution:** 1. Draw a line segment XY of length 8 cm. 2. At point X, use a protractor to draw an angle of \(45^ ext{o}\). Extend the line. 3. At point Y, use a protractor to draw an angle of \(70^ ext{o}\). Extend the line. 4. The intersection point of the two extended lines is Z. Triangle XYZ is formed. ### Exercises Try to construct the following triangles using a ruler, compass, and protractor where necessary: 1. Construct a triangle DEF with DE = 6 cm, EF = 7 cm, and FD = 8 cm. 2. Construct a triangle LMN with LM = 5 cm, MN = 7 cm, and angle LMN = \(80^ ext{o}\). 3. Construct a triangle GHI with GH = 9 cm, angle HGI = \(65^ ext{o}\), and angle GHI = \(35^ ext{o}\). 4. Construct a right-angled triangle JKL where the right angle is at K, JL (hypotenuse) = 10 cm, and JK = 6 cm. ### Assignment 1. Draw a triangle PQR where PQ = 7.5 cm, QR = 6.5 cm, and PR = 8.5 cm. Measure all the angles of the constructed triangle. 2. Construct a triangle ABC with AB = 7 cm, AC = 5 cm, and angle BAC = \(110^ ext{o}\). Measure the length of BC and the other two angles. 3. A plot of land is triangular in shape with one side measuring 100 meters. The angles at the ends of this side are \(50^ ext{o}\) and \(60^ ext{o}\). Draw an accurate scale diagram of the plot using a scale of 1 cm to 10 meters. Measure the lengths of the other two sides from your drawing. 4. Construct a right-angled triangle XYZ where angle Y is \(90^ ext{o}\), XZ = 13 cm, and XY = 5 cm. What is the length of YZ? (You can use the Pythagorean theorem \(a^2 + b^2 = c^2\) to check your construction). ### Summary In this lesson, we explored the fundamental methods for constructing triangles accurately using basic geometric tools. We covered four main cases: * **SSS (Side-Side-Side):** Used when all three side lengths are known. We use a compass to draw arcs from the endpoints of the base. * **SAS (Side-Angle-Side):** Used when two sides and their included angle are known. We use a protractor for the angle and a ruler for the sides. * **ASA (Angle-Side-Angle):** Used when two angles and their included side are known. We use a protractor for the angles and extend lines to find the third vertex. * **RHS (Right-angle-Hypotenuse-Side):** A specific case for right-angled triangles, requiring the hypotenuse and one other side. We construct a right angle and use a compass for the hypotenuse length. These construction skills are vital for solving various geometrical problems and have practical applications in many fields. Always ensure precision when using your tools for accurate constructions. ## , Construction of special angles: 60°, 30°, 90° and 45° ### Introduction Welcome to this lesson on geometrical constructions! In mathematics, constructions involve drawing geometric figures accurately using only two tools: a straightedge (an unmarked ruler) and a pair of compasses. You will learn how to construct specific angles such as 60°, 30°, 90°, and 45° without using a protractor. Mastering these fundamental constructions is crucial as they form the basis for constructing many other complex angles and geometric shapes. ### Key Concepts #### Tools for Construction * **Straightedge (Ruler):** Used to draw straight lines and connect points. It has no markings for measurement. * **Pair of Compasses:** Used to draw arcs and circles, and to transfer measurements. #### Construction of a 60° Angle This construction is based on the properties of an equilateral triangle, where all angles are \(60^\circ\). 1. Draw a straight line segment, say \(AB\). 2. With point \(A\) as the centre and any convenient radius, draw an arc that intersects \(AB\) at point \(P\). 3. With \(P\) as the centre and the *same radius*, draw another arc that intersects the first arc at point \(Q\). 4. Draw a straight line from \(A\) through \(Q\). The angle \(\angle QAB\) is \(60^\circ\). #### Construction of a 30° Angle To construct a \(30^\circ\) angle, you first construct a \(60^\circ\) angle and then bisect it. 1. Construct a \(60^\circ\) angle, say \(\angle QAB\), as described above. 2. With \(P\) (the intersection point on \(AB\)) as the centre and any convenient radius, draw an arc. 3. With \(Q\) (the intersection point on the first arc) as the centre and the *same radius*, draw another arc that intersects the arc drawn in step 2. Let this intersection point be \(R\). 4. Draw a straight line from \(A\) through \(R\). The angle \(\angle RAB\) is \(30^\circ\). #### Construction of a 90° Angle This construction creates a perpendicular line segment. There are a few methods; here we'll cover a common one. 1. Draw a straight line and mark a point \(P\) on it. 2. With \(P\) as the centre and any convenient radius, draw arcs that intersect the line on both sides of \(P\). Let these points be \(A\) and \(B\). 3. Now, with \(A\) as the centre and a radius *greater than \(AP\)*, draw an arc above (or below) the line. 4. With \(B\) as the centre and the *same radius*, draw another arc that intersects the arc drawn in step 3. Let this intersection point be \(Q\). 5. Draw a straight line from \(P\) through \(Q\). The angle \(\angle QPB\) (or \(\angle QPA\)) is \(90^\circ\). #### Construction of a 45° Angle To construct a \(45^\circ\) angle, you first construct a \(90^\circ\) angle and then bisect it. 1. Construct a \(90^\circ\) angle, say \(\angle QPB\), as described above. Let the point \(P\) be the vertex and \(PB\) be one arm. 2. With \(P\) as the centre and any convenient radius, draw an arc that intersects \(PQ\) at \(R\) and \(PB\) at \(S\). 3. With \(R\) as the centre and any convenient radius, draw an arc. 4. With \(S\) as the centre and the *same radius*, draw another arc that intersects the arc drawn in step 3. Let this intersection point be \(T\). 5. Draw a straight line from \(P\) through \(T\). The angle \(\angle TPB\) is \(45^\circ\). ### Vocabulary * **Construction:** The process of drawing geometric figures using only a straightedge and compasses. * **Straightedge:** An unmarked ruler used to draw straight lines. * **Compasses:** A tool used to draw circles or arcs and transfer distances. * **Angle Bisection:** The process of dividing an angle into two equal angles. If an angle has a measure of \(\theta\), its bisector creates two angles of \(\frac{\theta}{2}\). * **Perpendicular Bisector:** A line that cuts another line segment into two equal parts at a \(90^\circ\) angle. * **Equilateral Triangle:** A triangle in which all three sides are of equal length, and all three internal angles are equal to \(60^\circ\). * **Right Angle:** An angle measuring exactly \(90^\circ\). ### Examples #### Example 1: Constructing a 60° Angle Follow the steps outlined in the 'Key Concepts' section to construct a \(60^\circ\) angle at point A on a given line segment. Ensure your arcs are clear and distinct. #### Example 2: Constructing a 90° Angle Draw a line \(MN\). Mark a point \(X\) on \(MN\). Construct a line perpendicular to \(MN\) at point \(X\), thus forming a \(90^\circ\) angle. Use the method of drawing arcs from \(X\) to get points \(A\) and \(B\) on the line, then creating intersecting arcs from \(A\) and \(B\) above \(X\). ### Exercises 1. Draw a line segment \(PQ\) of length \(7 \text{ cm}\). Construct an angle of \(60^\circ\) at point \(P\) such that the angle opens upwards from \(PQ\). 2. Construct a \(90^\circ\) angle at point \(C\) on a straight line \(AB\). 3. From the \(60^\circ\) angle constructed in exercise 1, bisect it to obtain a \(30^\circ\) angle. 4. From the \(90^\circ\) angle constructed in exercise 2, bisect it to obtain a \(45^\circ\) angle. 5. Draw an arbitrary line segment \(XY\). Construct a line \(YZ\) such that \(\angle XYZ = 45^\circ\). ### Assignment 1. Construct an angle of \(15^\circ\). (Hint: How can you use the \(30^\circ\) angle?) 2. Construct an angle of \(75^\circ\). (Hint: Consider combining a \(90^\circ\) and a \(60^\circ\) angle, or constructing \(90^\circ\) and adding \(15^\circ\) or \(60^\circ\) and adding \(15^\circ\).) 3. Draw a line segment \(AB\) of length \(6 \text{ cm}\). Construct an equilateral triangle \(ABC\) using only a straightedge and compasses. 4. Construct a square \(DEFG\) with sides of \(5 \text{ cm}\) using only a straightedge and compasses. ### Summary In this lesson, you have learned the fundamental skills of geometrical construction using only a straightedge and a pair of compasses. You can now accurately construct angles of \(60^\circ\), \(30^\circ\), \(90^\circ\), and \(45^\circ\). These special angles are derived from basic geometric principles like the properties of equilateral triangles and angle bisection. Mastering these constructions is essential for further work in geometry, allowing you to create complex figures and angles by combining these basic techniques. Remember, precision in drawing arcs and lines is key to accurate constructions. ## Constructing the circumcircle of a triangle and ### Introduction Welcome to this lesson on geometrical constructions! Today, we will learn how to construct the circumcircle of a triangle. A circumcircle is a special circle that passes through all three vertices of a triangle. Understanding its construction is fundamental in geometry and has applications in various fields. ### Key Concepts #### What is a Circumcircle? A **circumcircle** of a triangle is the circle that passes through all three vertices of the triangle. The triangle is said to be inscribed in the circle. #### Circumcenter and Circumradius * The **circumcenter** is the center of the circumcircle. It is equidistant from all three vertices of the triangle. This means if O is the circumcenter and A, B, C are the vertices, then \( OA = OB = OC \). * The **circumradius** is the radius of the circumcircle, which is the distance from the circumcenter to any of the triangle's vertices. #### Perpendicular Bisector A **perpendicular bisector** of a line segment is a line that cuts the segment into two equal parts and is perpendicular to it. A key property is that any point on the perpendicular bisector is equidistant from the endpoints of the segment. #### Construction of a Perpendicular Bisector To construct the perpendicular bisector of a line segment AB: 1. Place the compass needle at point A and open it to a radius greater than half the length of AB. 2. Draw arcs above and below the line segment. 3. Without changing the compass radius, place the needle at point B and draw arcs that intersect the first two arcs. 4. Draw a straight line connecting the two points of intersection. This line is the perpendicular bisector of AB. #### Constructing the Circumcircle of a Triangle The circumcenter of a triangle is the point where the perpendicular bisectors of all three sides of the triangle intersect. This point is called the point of concurrency of the perpendicular bisectors. Follow these steps to construct the circumcircle of any triangle ABC: 1. **Draw the triangle:** Start by drawing the given triangle ABC. 2. **Construct the perpendicular bisector of side AB:** Use the method described above to construct the perpendicular bisector of side AB. 3. **Construct the perpendicular bisector of side BC:** Similarly, construct the perpendicular bisector of side BC. 4. **Locate the Circumcenter (O):** The point where these two perpendicular bisectors intersect is the circumcenter of the triangle. Label this point O. 5. **Check for the third bisector (Optional but recommended):** If you construct the perpendicular bisector of the third side (AC), you will find that it also passes through point O, confirming your circumcenter. 6. **Draw the Circumcircle:** Place the compass needle at the circumcenter O and open it to any one of the triangle's vertices (e.g., A, B, or C). The distance \( OA \) (or \( OB \) or \( OC \)) is the circumradius. Draw the circle. This circle will pass through all three vertices of the triangle. ### Vocabulary * **Circumcircle:** A circle that passes through all three vertices of a triangle. * **Circumcenter:** The center of the circumcircle; the point of concurrency of the perpendicular bisectors of the sides of a triangle. * **Circumradius:** The radius of the circumcircle; the distance from the circumcenter to any vertex of the triangle. * **Perpendicular Bisector:** A line that intersects a line segment at its midpoint at a \( 90^ ext{o} \) angle. * **Equidistant:** At equal distances. * **Concurrent lines:** Three or more lines that intersect at a single common point. ### Examples #### Example 1: Constructing the circumcircle for an acute-angled triangle Let's construct the circumcircle for an acute-angled triangle PQR. 1. Draw an acute-angled triangle PQR. 2. Construct the perpendicular bisector of side PQ. Label it \( L_1 \). 3. Construct the perpendicular bisector of side QR. Label it \( L_2 \). 4. The intersection of \( L_1 \) and \( L_2 \) is the circumcenter, O. 5. Place the compass at O, open it to P (or Q or R), and draw the circle. You will observe that the circumcenter O lies *inside* the triangle for an acute-angled triangle. #### Example 2: Locating the circumcenter for different types of triangles * **Acute-angled triangle:** The circumcenter lies **inside** the triangle. * **Right-angled triangle:** The circumcenter lies **on the midpoint of the hypotenuse**. * **Obtuse-angled triangle:** The circumcenter lies **outside** the triangle. ### Exercises 1. Draw an equilateral triangle with sides of 6 cm. Construct its circumcircle. What do you notice about the position of the circumcenter relative to the triangle? 2. Draw a right-angled triangle with sides 3 cm, 4 cm, and 5 cm. Construct its circumcircle. Where does the circumcenter lie in relation to the triangle and its sides? 3. Draw an obtuse-angled triangle ABC. Construct its circumcircle. Where is the circumcenter located for this type of triangle? 4. Explain why the circumcenter is equidistant from all three vertices of a triangle. ### Assignment 1. Construct a triangle with side lengths 7 cm, 8 cm, and 10 cm. Then, construct its circumcircle. Measure the circumradius. 2. Using a ruler and a pair of compasses, construct a triangle XYZ where \( XY = 6 ext{ cm} \), \( YZ = 8 ext{ cm} \), and \( ext{angle } XYZ = 90^ ext{o} \). Construct the circumcircle of triangle XYZ. What is the length of the circumradius? 3. Research and explain a real-world application where the concept of a circumcircle or circumcenter is used. ### Summary Today, we learned about the circumcircle of a triangle, which is a circle passing through all three vertices. We discovered that the center of this circle, called the circumcenter, is found at the intersection of the perpendicular bisectors of the triangle's sides. The distance from the circumcenter to any vertex is the circumradius. We also noted that the position of the circumcenter varies depending on the type of triangle: inside for acute, on the hypotenuse for right-angled, and outside for obtuse triangles. Mastering this construction is a vital skill in geometry.