# Thread: [SOLVED] construction of equilateral triangle

1. ## [SOLVED] construction of equilateral triangle

Can someone help me with this:
construct equilateral triangle if vertex A is given and vertices B and C belong to given lines b i c.

Thanks.

2. If vertex A is given, then you have it's co-ordinates.

For B, label the point $(x_2,y_2)$

For C, label the point $(x_3,y_3)$

now use the line equations to write one variable in terms of the other
in both cases.

Then use Pythagoras' theorem since the distance between A and B
equals the distance from A to C equals the distance from B to C.

3. I can't use coordinate system so i don't have coordinates of vertex A.I think that isometric transformations(rotation) have to be used to solve this but i'm not sure.

4. Yes,
sorry i thought the problem involved given the line equations and point A.

5. Does anyone know how to solve this?

6. The circles are for construction of isosceles triangles.
Their common centre is the given point.
An equilateral triangle is a special case of isosceles triangle.
Hence the use of the purple line with a 120 degree angle as shown.
We only need a parallel line to this from the point to find one triangle side.
the blue parallel lines find the other two.

7. Why do you need two blue lines and circles?

8. The blue lines are parallel,
the red line cuts all the blue segments between the green given lines at their midpoints.
i use the circles since every point on the circle is equidistant from the given point,
therefore the diagram is creating isosceles triangles with the given point as vertex.
Only one circle is needed really, i use 2 for emphasis.
The problem is to find the isosceles triangle that is also equilateral.
Hence, the purple line is used to locate the 120 degree angle,
meaning the inner angle is 60 degrees.
Then the parallel dotted lines discover an isosceles triangle with two 60 degree angles,
hence the 3rd angle is also 60 degrees, therefore it's equilateral.
That is my thinking on it.

9. Can you prove that?Check this image, it shows that your construction isn't OK.

10. Your lower dotted purple line should be adjusted to intersect the green
and blue lines, giving three 60 degree angles.

i have to go for a few hours,
but i will double-check everything later.

11. My diagram only works if the point is on the bisector of the angle between the 2 given lines.
I'll try to find the solution to the general case.

12. Let A' be the reflection of A in the line c. Draw a line through A' making an angle 30º with AA', meeting the line b at Y. Let X be the midpoint of AY. Draw a line through X, perpendicular to AY, meeting the line c at Z. Then the triangle AYZ is equilateral.

Reason: the construction ensures that the triangles AA'Z and AYZ are isosceles, so that A'Z = AZ = YZ. Thus the points A', A, Y all lie on a circle with centre Z. Thus angle AZY is 60º (angle at centre = twice angle at circumference). It follows that the isosceles triangle AZY is equilateral.

There is a second equilateral triangle with a vertex at A and the other vertices on b and c, obtained by drawing the line A'Y at an angle of 30º on the other side of AA' (to the left of AA' instead of to the right as in my picture).

13. That is excellent.Thanks

14. If the angle between the lines is bisected and the perpendicular bisector drawn, followed by drawing the perpendicular to this from the given point
then we can send 2 lines at 30 degrees to the bisector, from the point of intersection of the 2 construction lines.
The green triangle is isosceles at the other end, so it is equilateral also.

Therefore, using the given point as the circle centre, we can draw the black circle, then use the given point as centre, draw a circle of the exact same radius.
This is the dashed circle.
Where this circle intersects the lines gives the other 2 triangle vertices.

15. Here's another view, using rotating equilateral triangles
and parallelograms.