Draw a Hexagon with Parallel Lines: 6-Sided Shape Drawing Homework

[Gary Smart]

Homework Statement

The problem involves drawing a particular set of lines and analysing anything of interest. The steps are as follows:
: Begin with a hexagon with standard notation.
: Draw a line (T) that passes through point B and which is parallel to AC.
: Produce the lines AD and EC, remember to jot X at the intersection point.
: Create line NX, with N on line (T) and NX is right angles to T.
: Produce a circle with XB as the diameter.

The Attempt at a Solution

I have attached a copy of my drawing, it doesn't have the construction lines because it is only rough, I'm firstly interested to see if I have the general idea.

Gary.

 

[mfb]

Looks good so far.

You could label T and mark the right angle at N.

 

[Campbell1982]

Looks good so far.

You could label T and mark the right angle at N.

Thank you for your help mfb. I have drawn the hexagon and have attached a copy. I have labelled T and marked the right angle at N. I'm interested also in finding anything interesting about the circle. The only thing I can see really is the number of triangles produced inside of it and that it is in contact with two vertices.

 

[mfb]

The circle seems to go through N and C, it would be interesting to see if this is a coincidence.
Does the hexagon have to be regular as in your second sketch?

 

[Campbell1982]

The circle seems to go through N and C, it would be interesting to see if this is a coincidence.
Does the hexagon have to be regular as in your second sketch?

The hexagon has to be regular.

 

[Gary Smart]

The circle seems to go through N and C, it would be interesting to see if this is a coincidence.
Does the hexagon have to be regular as in your second sketch?

Triangle BXN because is circumscribed. XB is a diameter and the midpoint of line XB is the centre of the circle. This also means that any line constructed from line XB will be a right angled triangle (circle theorem: angles in a semi-circle). Also, I think the midpoint between A and the hexagon's centre will produce a cyclic quadrilateral. I can't find anything else of particular interest.

 

[mfb]

I don't understand some parts of the description but the right angle is the important part (also for C).

I can't find anything else of particular interest.

Me neither.

 

[Gary Smart]

Would you agree that any line produced from line XB to the the circle's diameter will produce a right-angled triangle? What do you mean about the right angle being important and C?
Do you think there is anything of interest to point out about Euler's line and the circle?

 

[mfb]

Would you agree that any line produced from line XB to the the circle's diameter will produce a right-angled triangle?

I don't understand which triangle you mean.

What do you mean about the right angle being important and C?

Without the right angle, the argument does not work.

Do you think there is anything of interest to point out about Euler's line and the circle?

What, where?

 

[Gary Smart]

I don't understand which triangle you mean.
Without the right angle, the argument does not work.
What, where?

I wrote that statement slightly wrong, what I actually meant was: Any triangle that has a point on the circumference of the circle with line XB (line XB is a diameter of the circumcircle) as one side will always produce a right-angled triangle.

What do you mean by the argument?

 

[mfb]

Any triangle that has a point on the circumference of the circle with line XB (line XB is a diameter of the circumcircle) as one side will always produce a right-angled triangle.

Sure, but how is that relevant?