Prove with Euclid's Elements Based on Book I of Euclid's Elements, up to Theorem 30:
Prove the following sentence, or give an example where you show that the sentence is false.
It is possible to build a triangle $\triangle ABC$ knowing the side $BC$  the altitude $AD$ with $D$ on the line $BC$ and the angle $\angle BAC=\alpha$ less than two right angle
I think how $A$, $B$ and $C$ are vertices and $AD$ is on $BC$ then $D=B$ or $D=C$ and how $AD$ is the altitude the the triangle has to be a right triangle and then it will be a right angle and  the angle $\alpha$ plus the angle remaining should be equal to one right angle then $\alpha$ is less than two right angles
 A: Thank you for showing your work.  I am having trouble understanding your work,
so I can't react to it.  However, I can give you my take on the problem.  Note
that I am speaking from a background of basic trignometry.  This means that
since I am unfamiliary with Euclid's Elements, it is unclear how valid my
answer is (with respect to Euclid's Elements).
Assume that line BC lies flat on the X axis, with D located at the origin (0,0).
If I interpret the problem correctly, point A can be presumed to be a point on
the positive side of the y axis.  Further, since the length of the altitude
AD is known, the precise location of point A is known.
Further, you know two things:
(1) The length of side BC 
(2) The measure of $\angle BAC = \alpha.$
I would approach this (complicated) trignometry problem as follows:
Let $\beta = \angle BAD, \gamma =  \angle DAC \Rightarrow (\beta + \gamma = \alpha)$.
You should be able to express the length of line BD as : 
(tangent of $\beta$) $\times$ (the length of line AD).
Similarly, you should be able to express the length of line DC as : 
(tangent of $\gamma$) $\times$ (the length of line AD).
Further, you know that length of BD + length of DC = the length of BC, which
is given as a known length.
My instinct is that this should be sufficient to completely determine the dimensions
of the triangle; however I am actually not 100% sure of this.
Anyway, this answer gives you a guide for how to attack the problem, assuming
that it is a problem from basic trignometry.
As I said, you also have to consider how this problem-attack strategy is affected
by the background of the problem being Euclid's Elements (again which I am unfamiliar with)
rather than basic trignometry.
Addendum
I think that it is worthwhile to also provide a (totally non-mathematical) intuitive approach to the problem.
If I understand correctly, the problem merely asks whether the information provided is sufficient to completely determine the triangle's dimensions.
Assume that $(2r) = $ the (known) length of line BC.
Further assume that point B is located at (-r,0) and point C is located at (r, 0).
Assume that the length of line segment AD, which is (presumably known), is given as $s$.
Now imagine the horizontal line given by the equation $y = s.$
Point A must be somewhere on this line.  Further the location of point A will determine the measure of $\angle BAC,$ which is apparently known to $= \alpha$.
As the possible locations of point A move across the line $y = s$, the resultant $\angle BAC$ changes.
Intuitively, it seems to me that there would be only one point A on line $y = s$ that generates an $\angle BAC = \alpha.$
Again, I am not 100% sure, but I (still) consider intuition a powerful weapon here.
Note, for reasons of symmetry, it seems reasonable to assume that the x coordinate of point A is $\geq 0.$
This is because if there is a satisfying point $A$ with coordinates $(-a, s)$ where $a$ is assumed to be positive,  then I would expect that alternatively locating point $A$  at $(a, s)$ would also satisfy the constraints.
In this event, it seems to me that the two locations for point A, $(-a, s)$ and $(a, s)$ would result in two triangles that are reflections of each other.  In that event, the question is whether these reflections are to be considered distinct triangles.
A: In this answer I'll use Prop. 21 in Book III of Euclid's Elements:

In a circle the angles in the same segment equal one another.

Hence if we construct a single triangle $ABC$ with $\angle BAC=\alpha$, for instance an isosceles triangle with base angles $(2\pi-\alpha)/2$, all points $A'$ of that segment of circle $ABC$ containing $A$ will make the same angle: $\angle BA'C=\alpha$.
We can then draw a line parallel to $BC$ on the same side as $A$, and at a distance from $BC$ equal to the given altitude of the triangle: if that line intersects at $E$ the segment of circle constructed above, then triangle $ABE$ is a solution. This happens if the given altitude is not greater than the height of the circular segment.
But if the given altitude is greater than the height of the circular segment, then no solution is possible.
A: Hint:  Sometimes its a matter of visualizing what you do and don't know.  You have points $B,C$ and line $BC$ and you know the angled line that $A$ must be on, and you know how far away $A$ must be vertically from the line $BC$.  So its a matter of proving/finding the one point that is both on the angled line, and is the distance vertically from $BC$.
So.... how can you do that?

Solution:

  Pick any point $E$ on $BC$.  Construct a line perpendicular to $BC$ at $E$ and mark of a point $F$ on that line that is the known distance of the given altitude $AD$.  Construction a line that is parallel to $BC$ at $F$.


  A parallel line consists of all the points that are the distance $AD$ from $BC$ so the point $A$ will be on that parallel line.  And $A$ must be on that angled line.


 And as the angled line is not parallel to $BC$ it is not parallel to the parallel line through $F$.  So the angled line and the line through $F$ and the angled line will intersect at a unique point.  That unique point must be $A$.

