Equilateral triangle ABC is inscribed in a circle

In summary, the conversation is about solving a geometry problem involving an equilateral triangle inscribed in a circle. The problem statement is to prove that the points A, D, and E lie on a straight line. The conversation includes discussions on using the Theorem of Inscribed Angles and setting up a coordinate system. The conversation also mentions finding the angles of the blue triangle and proving that ADE is a straight line. The conversation ends with a reminder to explain statements while solving a problem.
  • #1
dawo0
5
0

Homework Statement


Can someone help me solve this, and teach me how to solve such problems in future? An equilateral triangle ABC is inscribed in a circle . Point D lies on a shorter arc of a circle BC. Point E is symmetrical the point B relating to the line CD . Prove that the points A, D , E lie on one straight line. If something is unclear just tell me.

Homework Equations


--------

The Attempt at a Solution



I do not know how to go about Himself .
Frame I figure .
 

Attachments

  • konkurss.png
    konkurss.png
    3.1 KB · Views: 591
Physics news on Phys.org
  • #2
Welcome to PF!
dawo0 said:

Homework Statement


Can someone help me solve this, and teach me how to solve such problems in future? An equilateral triangle ABC is inscribed in a circle . Point D lies on a shorter arc of a circle BC. Point E is symmetrical the point B relating to the line CD . Prove that the points A, D , E lie on one straight line. If something is unclear just tell me.

.
Use the Theorem of Inscribed Angles.
 
  • #3
Excellent English! Much better than before. Congratulations. This may be more than is necessary but I am more comfortable with "analysis" than "geometry" so I would set up a coordinate system with origin at the center of the circle and (0, 1) at point C. Then the equation of the circle, in that coordinate system, is [itex]x^2+ y^2= 1[/itex]. On can show that the line through the point A and the center of the circle is [itex]y= x/2[/itex] so that point A has coordinates where that line intersects the circle. With [itex]y= x/2[/itex] the equation of the circle becomes [itex]x^2+ x^2/4= (5/4)x^2= 1[/itex] so that [itex]x= -\frac{2}{\sqrt{5}}= -\frac{2\sqrt{5}}{5}[/itex] and [itex]y= -\frac{\sqrt{5}}{5}[/itex]. Similarly for point B. With that information it should be fairly easy to find the coordinates of points D and E and show that they line on a single line.
 
  • #4
I figured all the angles. What next?
 

Attachments

  • Kąty.png
    Kąty.png
    16 KB · Views: 560
  • #5
Try to determine the angle ADE. (First find the angle of the blue right triangle at D.

circleangles.JPG
 
  • #6
How do I determine the angle ADE as it is simple? .
Calculated the angles of the blue triangle.
Help ! "D l
 
  • #7
You have to prove that ADE is a straight line, that is, the angle ADE is 180°. What are the angles ABC, ADC ADB,?
The red line CD halves the blue triangle, as it is symmetric to the red line. Wjhat are the angles at D?
 
  • Like
Likes dawo0
  • #8
All I calculated.
I can't write request.
 

Attachments

  • 29 listopad 2014 (2).gif
    29 listopad 2014 (2).gif
    16.8 KB · Views: 783
  • #9
dawo0 said:
All I calculated.
I can't write request.

Show what you calculated. What are the angles γ and δ; β and λ? How many degrees? What is the angle σ?

circleangles.JPG
 
  • #10
Already pasted.
circleangles-jpg.75935.jpg
29-listopad-2014-2-gif.75931.gif
 

Attachments

  • circleangles.JPG
    circleangles.JPG
    10.6 KB · Views: 615
  • #11
You seem to be assuming that the line drawn from A will intersect BC at a 90 degree angle. This is incorrect.
In post #2, ehild said to use the law of inscribed angles. Check that out and see if you can find all the angles in terms of the one variable angle.
 
  • #12
@dawo0: your drawing does not help to solve the problem. It shows the special case when the point D is at the middle of the arc BC. But you need to prove the statement of the problem for any point D on the arc.
You have drawn ADE as a straight line, but you have to prove that it is a straight line!
Do you understand why are γ=δ=60°and λ=β=60°? You should explain. In general, you should explain your statements when solving a problem.
Something like that: β is one angle of the equilateral triangle, so it is 60°. β and λ are inscribed angles of the circle, belonging to the same arc AC, so λ=β...

29-listopad-2014-2-gif.75931.gif
[/QUOTE]
 

FAQ: Equilateral triangle ABC is inscribed in a circle

What is an inscribed equilateral triangle?

An inscribed equilateral triangle is a triangle that is drawn inside a circle, with all three vertices touching the circle. This means that the sides of the triangle are tangent to the circle, and the triangle's angles are all equal to 60 degrees.

How is an equilateral triangle inscribed in a circle?

An equilateral triangle can be inscribed in a circle by drawing three lines from the center of the circle to each of the triangle's vertices. The point where these lines intersect is the center of the circle, and the triangle's sides will be tangent to the circle at these points.

What is the relationship between an inscribed equilateral triangle and its circumscribed circle?

The circumscribed circle of an inscribed equilateral triangle is the smallest possible circle that can contain the triangle, with all three vertices touching the circle. This means that the radius of the circle is equal to the length of any side of the triangle.

How can the properties of an inscribed equilateral triangle be used in geometry problems?

The properties of an inscribed equilateral triangle, such as the fact that its angles are all equal to 60 degrees and its sides are tangent to the circle, can be used to solve various geometry problems. For example, they can be used to find the area or perimeter of the triangle, or to find the length of a side or angle.

What are some real-life applications of an inscribed equilateral triangle?

An inscribed equilateral triangle can be seen in various structures, such as bridges, towers, and arches, where it is used to distribute weight evenly and provide stability. It is also commonly used in geometry and trigonometry problems, as well as in art and design.

Similar threads

Back
Top