Triangle inscribed in a circle

[Kolika28]

Homework Statement

[/B]
In a circle with center S, DB is the diameter. The line AC goes 90 degrees from the center point M of the line SB. "
What type of triangle is ACD?


2. Homework Equations

The Attempt at a Solution

I can see it is an equilateral triangle, but do not know how to explain it. I know it has something to do with the center of the circle S. I believe it may have something to do with the line SM as it goes through the midpoint of the side AS. But I do not know what to do after this, I'm really lost.
I really appreciate some help :)

 

[Orodruin]

Please give the entire question exactly as stated. Also provide your own attempt and what ideas you have in more detail.

 

[Kolika28]

Ok, I will edit the question

 

[Helly123]

Which line has the same length?

 

[Kolika28]

I have also been thinking that since the line AC goes 90 degrees through the center point M on SB and therefor also line DB , AD=DC because point A and C have the same distance from point D. But then again, how can I prove that AC is equal to the other sides?

 

[Orodruin]

I can see it is an equilateral triangle

Never base your reasoning on how a drawing looks. It can easily mislead you. I am not saying that your conclusion here is wrong, just do not trust figures. Let the maths speak for itself.

AD=DC because point A and C have the same distance from point D

This much is true and it follows directly from symmetry. So you have at least shown that the triangle is an isosceles triangle. It remains to find the distance of AC, or the distance AM, which will give you the distance AC. What can you say about the distances SM and SA? How do they relate to the distance AM?

 

[Enelafoxa]

Ups.. Sorry.. I am new here.. Wehre i can Read the guideline?

 

[fresh_42]

Ups.. Sorry.. I am new here.. Wehre i can Read the guideline?

https://www.physicsforums.com/threads/physics-forums-global-guidelines.414380/

 

[Helly123]

I have also been thinking that since the line AC goes 90 degrees through the center point M on SB and therefor also line DB , AD=DC because point A and C have the same distance from point D. But then again, how can I prove that AC is equal to the other sides?

for my method is that first find a proof that AC AD CD are the same length. how to proof that? I think the S point can give a clue/

 

[Chestermiller]

If SB is equal to the radius r, in terms of r, what is the length of SM? What is the sine of angle SAM?

 

[Kolika28]

Chestmiller, the sine of the angle SAM will then be 0.5r/r?

 

[Chestermiller]

Chestmiller, the sine of the angle SAM will then be 0.5r/r?

Can't you cancel the r's? What angle has this value for its sine?

 

[Kolika28]

The value will then be 0.5 and the angle SAM must be 30 degrees. Which means that the triangle AMS is as 30,60,90 triangle.

 

[Chestermiller]

Can you prove that triangles SAM and DAM are similar triangles?

 

[Kolika28]

Both triangels share the same angle M and the side AM.

 

[Chestermiller]

Both triangels share the same angle M and the side AM.

That is not sufficient to prove that they are similar. You need to show that all three angles are equal.

 

[Kolika28]

I'm sorry, but I'm lost right now.