|We first prove that ∆BCA is a right triangle
||m∠BCA = 90°
||BC was constructed using the procedure in
Perpendicular to a line at a point.
See that page for proof.
||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
||AC = the given leg
||AC was copied from the leg at the same compass width
|Now prove ∠BAC is the given angle A
||m∠BAC = given m∠A
||Copied using the procedure in Copying an angle.
See that page for proof
||∆BCA is a right triangle with the desired hypotenuse H and angle A
||From (2), (3), (4)