We first prove that ∆BCA is a right triangle 
1 
m∠BCA = 90° 
BC was constructed using the procedure in
Perpendicular to a line at a point.
See that page for proof.

2 
Therefore ∆BCA is a right triangle 
By definition of a right triangle, one angle must be 90° 
Now prove AC is congruent to the given leg 
3 
AC = the given leg 
AC was copied from the leg at the same compass width 
Now prove ∠BAC is the given angle A 
4 
m∠BAC = given m∠A 
Copied using the procedure in Copying an angle.
See that page for proof 
9 
∆BCA is a right triangle with the desired hypotenuse H and angle A 
From (2), (3), (4) 