# Angles with Specific Conditions

omairsiddiquer shared this question 5 months ago

In △ABC, (∠B > ∠C), I want to draw two distinct points P and Q on AC such that ∠PBA = ∠QBA - ∠ACB. How can I apply this condition to my diagram, so that whenever I move one of the vertices of △ABC, the condition ∠PBA = ∠QBA - ∠ACB is always satisfied? 1

Attached is an example. The point P can be moved along the segment AC, the point Q on the line through A and C is placed automatically. (Of course Q might land on the outside of the triangle if the ∠PBA + ∠ACB happens to be greater than ∠CBA.) 1

Para cada Q hay todo un lugar geométrico para P

creo que la pregunta tienen tanta libertad que la respuesta no es nada útil salvo que los posibles P están situados simétricamente respecto la recta BA

Files: foro.ggb 1

@mathmagic Did you overlook that P and Q are supposed to be on the segment (or line?) AC?

It's true that the question is somewhat ambiguous but not as much as you thought. I one of the points P or Q is fixed, then the other is uniquely determined. If the plan was to allow moving P freely, then the solution is as simple as I provided. (Similarly, if Q is freely movable and P is computed.)

Of course, there might be an unmentioned additional condition (for example symmetry on AC, i.e., |AP|=|CQ|), in which case the solution would have to be more sophisticated. So let's see what omairsiddiquer actually wants. 1 