Assuming I'm understanding the problem statement correctly and we have a vertical load transferred to a beam via a sort of angled frame, then here's how I'd think about the beam reactions.
The Quick Way...
The quickest approach is to solve for the support reactions by remembering the structure must be globally stable. In this case, we can consider the beam and the frame as a single structure for purposes of calculating the support reactions. Our equations of statics say the sum of the forces in the horizontal direction, the sum of the force in the vertical direction, and sum of the moments, must each be zero. Because there is no externally-applied horizontal load, there is only one possible horizontal force (the support reaction). For the sum of horizontal forces to be zero, the horizontal support reaction must therefore also be zero.
The Longer Way...
The longer way round is perhaps to convince ourselves that the frame applies equal and opposite horizontal loads to the beam and that this, again, means the horizontal support reaction must be zero.
First, imagine we have a simply supported beam with equal and opposite horizontal load applied at a point.
There is no net load at the point, so it's reasonably easy to convince ourselves that there's no support reaction.
What if we applied equal and opposite loads at separate points along the beam?
Now, it's perhaps less clear that there's no support reaction. But we can convince ourselves easily using the equations of statics. The beam reactions are about global stability of the structure. As is always the case in statics, the sum of horizontal forces must be zero.
$$\Sigma F_x = (+P) + (-P) + H_A = 0$$
So, by some quick math, we prove that the horizontal reaction at Point A must be zero. In fact, for this beam configuration, the only segment of the beam "feeling" axial load is the segment in between the applied loads. It can be tempted to picture the left load "pushing" on the beam segment to the left of it, but that's not what's happening. Imagine a game of tug-of-war where the two innermost opponents perfectly oppose each other's forces...the players farther down the rope wouldn't feel anything, right?
So now, all we have to do is convince ourselves the horizontal components of the applied loads are equal and opposite.
If we zoomed in and just looked at the bit of structure right were the vertical load is applied...well, that bit of structure must also be in equilibrium. The applied load is going to distribute between the two angled members, which means there will be horizontal components. However, again, by our lovely equations of statics, for this bit of structure to be in equilibrium, the sum of horizontal forces must be zero. Here, the only possible horizontal forces are the horizontal components of the loads in the angled members -- and thus, these components must be equal and opposite.
Ta, da! Sometimes our initial intuition can lead us a little astray but the equations of statics never lie. Forces and moments must always sum to zero in order for the structure to remain...static.