Complete a triangle given orthocenter, 1 vertex, two lines each containing a vertex

[mathpanther]

Hello,
I've been racking my brain trying to figure this one out using trigonometric or vector math. I'm no math expert and this is over my head. I found this site and am hoping I can get some help.

I'm trying to figure out an approach to solve a triangle given the following information. I've attached an image to help clarify the given information:
- The Orthocenter (P)
- 1 Vertex (A)
- 2 lines pointing towards the two unknown vertices (Line DE and Line FG)
From this, I need to find points B and C



Since I have the orthocenter (P) and the one vertex (A), I know the distance from vertex A and the orthocenter P. I know the opposite side of the triangle is perpendicular to Line AP and vertices B and C are where the opposite side crosses the lines DE and FG. All I need to solve the triangle is the distance from the known vertex to the opposite side. There's too many moving targets for me to wrap my head around.

What I'm hoping to find are formulas that I can plug the given data into that will help me find the triangle. Any ideas on how to approach this?

Best regards

 

[John_MathHelpMan]

Orthocenter question

It's very late and I'm about to hit the sack, so I don't have time to give this the thought it deserves. Having said that, it may well be that a path to the solution might be found in the fact that if you have three vertices and the orthocenter, any one of the four points is the orthocenter for the other three. If you haven't come up with something by tomorrow, I'll put some more time in on it.

 

[mathpanther]

Updated original post

I've updated my original post with a labeled graphic and tried to clarify a bit. Please see the original post.
Thanks.

 

[mathpanther]

It's very late and I'm about to hit the sack, so I don't have time to give this the thought it deserves. Having said that, it may well be that a path to the solution might be found in the fact that if you have three vertices and the orthocenter, any one of the four points is the orthocenter for the other three. If you haven't come up with something by tomorrow, I'll put some more time in on it.

Thank you John. That's one piece of information I did not consider. The problem here is that I only have two of the four points. I think the key is to somehow find the distance from the known vertex (A) to where it intersects perpendicularly with the opposite side (BC).

Thanks for your help.

 

[John_MathHelpMan]

Orthocenter Question

Please share what you know about the one vertex and the orthocenter as well as how the two vectors DE and FG are defined. I'll take some time today to play with it.

 

[mathpanther]

Please share what you know about the one vertex and the orthocenter as well as how the two vectors DE and FG are defined. I'll take some time today to play with it.

Hi John,
Thanks so much for looking into this.

The one vertex (A) and orthocenter (P) are arbitrary points, as are vectors DE and FG. If it helps, I'll give you the actual point locations (rounded to 4 decimal places) in my diagram:
Orthocenter (P): 0, 0
Point locations:
A: 2.5463, -1.5345
D: -0.3026, -0.5389
E: -1.4161, -0.7023
F: -1.147, 1.2116
G: -0.5948, 3.3767

Completing the triangle graphically in my CAD software, gives me these answers:
B: -4.6337, -1.1745
C: 0.3563, 7.1055
Distance from A to P: 2.9729
Distance from A to the perpendicular intersection of line BC: 6.3354
The perpendicular intersection of the line from A to the line BC: -2.8799, 1.7356

Does this help any?

 

[mathpanther]

It looks like one solution is not always possible

I found a great App for the Mac called GeoGebra that allowed me to setup the needed constrains, move line BC, and interactively see when the solution triangle is met. The only problem is that certain configurations of lines DE and FG can create cases where the solution is never met or where the solution is met with two different triangles. I would need to investigate how to constrain things so that only one solution is found. Rats! I need to take a break from this and get some real work done. I have to look at it a little later.

Thank you John, and anyone else that has thought about this!

Best regards