How to draw an equilateral triangle on a circle

The steps are as follows;

Take a little ace on the circle

Follow the straight line where A and Center B are located.

Fold the circle in half, and the intersection point of the polyline and the circle is o (the first point).

Fold points A and B in half, so that points A and B coincide, the polyline is perpendicular to the diameter ao, and the polyline and the circle have two intersection points, P and Q respectively

Connect OP, OQ, and PQ to get an equilateral triangle OPQ

Extended Materials. The concept and properties of equilateral triangles.

C is the moving point on the line segment AE (not coincident with A and E), and the regular triangle ABC and regular triangle CDE are made on the same side of AE, where AD and BE intersect at the point O, AD and BC at the point P, and BE and Cd at the point Q.

1、ad=be；2、ap=bq；3、de=dp；4、∠aob=60°。What are the conclusions of Heng Zheng?

Analysis: 1. This is a very classic common question with one picture and multiple conclusions. To prove AD=BE, ACD BCE needs to be proved.

The conditions for the congruence of these two triangles are AC=BC (the triangle ABC is an equilateral triangle) and CD=CE (the triangle CDE is an equilateral triangle).

Since acb=dce=60° (all three inner angles of an equilateral triangle are 60°), DCB=180°-60°-60°=60°.

2. Because acd= acb+ dcb=60°+60°=120°, bce= dbc+ dce=60°+60°=120°, so acd= bce. ACD BCE can be demonstrated at this point.

3. ACD BCE can be concluded: AD=BE (conclusion 1 is true), cap=CBQ, i.e. 1= 2. You can take advantage of the "corner corners":

1= 2,ac=bc, acp= bcq=60° proves acp bcq; So ap=bq (conclusion 2 is true), cp=cq.

Starting from a certain point in the circumference of the circle, the arc is drawn with the radius of the circle as the radius in turn, and a total of 6 intersection points are obtained, and these six intersection points are connected by selecting one after another, which is an equilateral triangle.

<>1. First draw a line segment with a straight ruler;

2. Then tie a compass at an end, take the length of the line segment with the radius, and then draw an arc;

3. Tie the compass to the other end, and then repeat to draw an arc;

4. Draw lines from the intersection of two arcs to the two endpoints;

5. Finally, the arc is erased, and the triangle composed of three line segments is an equilateral triangle.

<> drawing an equilateral triangle requires the following steps:

1.Use a straight jujube ruler and pencil to draw a straight line segment as the base edge of the equilateral triangle.

2.Use a ruler to draw a line segment perpendicular to the bottom edge of the bottom edge at the bottom edge, such as a bend, as the high line of an equilateral triangle.

3.Use a ruler to draw two diagonal segments on either side of the high line, each at an angle of 60 degrees to the bottom edge.

4.The line segments that connect the two sides form an inverted triangle with the original bottom edge.

5.Check that the three sides of the triangle are equal. If the edge lengths are different, you can use a ruler to measure and correct the edges.

6.Make sure that all the line segments meet the requirements, and then use a pencil to outline the contour of the entire three-slag rock sullen shape.

7.Use colored pens or paints to fill in the triangle enclosed area to reveal a clear pattern of equilateral triangle edges.

It is important to note that the tools used to draw equilateral triangles are best rulers and pencils, and use clear straight lines and angles to ensure that the edges of the triangle are clear and compliant. In addition, you can also use computer graphics software to draw equilateral triangles, which is more convenient and fast.

The first step is to draw one side of the equilateral triangle you want to draw (line segment ab). The second step, the two ends of the line segment you draw are the center of the circle, the length of the line segment is the radius to draw the circle, the intersection point c of the two circles is the third vertex of the equilateral triangle, and the two ends of the line segment drawn in the first step are drawn together to draw an equilateral triangle ABC.

Method 1: Draw an equilateral triangle first, then draw the center of gravity of the triangle (the difference between the three high lines, the angle bisector, and the focus of the middle line) as the center of the circle, and draw a circle to the radius at the vertex of the triangle.

Method 2: Draw a circle first, and then divide the circle into three equal parts, (the center angle of the circle is 120 degrees) The three points intersected by the virtual and bright are the vertices of the triangle.

1. Method 1:

Connecting the center of gravity with the three vertices gives you three congruent triangles.

The center of gravity of the triangle is the intersection of the midlines of the three sides of the triangle. When the geometry is a homogeneous object, the center of gravity coincides with the centroid. ）

2. Method 2:

Divide any side into thirds, and connect the equal points with the opposite vertices to get three triangles with equal bases and the same height.

3. Method 3:

Connecting the center of gravity with the midpoint of the three sides gives three quadrilaterals of lead congruence.

Draw a standard equilateral triangle step from a ruler:

1. Parallel lines of unit width can be made along both sides of the ruler, and then a straight line can be drawn casually.

2. Then use translational compound theorems to extend ab to c, where bc=ab. Then we use the perpendicular theorem for the perpendicular line, so that we get the right angle, and then we use the Pythagorean theorem.

3. Use the rotary replication theorem to turn DC down to DE, then connect CE, and then use the rotary replication theorem to turn CE up to become CF, and then use the translational replication theorem to move down to become DG.

4. Then do it again to get the line segment DH, and then use the rotation complex theorem to turn it down to become di, and the triangle CIE is an equilateral triangle.

Draw a line segment 1 (e.g., 1cm) of known length, take a certain endpoint of the line segment as the axis, use a protractor to measure the 120° square shed burning direction clockwise with line segment 1 as the starting edge, draw the line segment 2 with the same length chain friend along the direction, measure the direction of 120° counterclockwise with the line segment spring 1 as the starting edge, and draw the line segment 3 of the same length along the direction. The vertices that connect the three segments are equilateral triangles.

Draw a straight line AB, with the two endpoints of the straight line AB in front of the center of the Hui base circle, and its length is a point C for the radius author and the arc, connecting AC and BC, forming an isosceles triangle ABC.