Step 2 drop perpendicular from [math]B[/math] to the x-axis and call that point [math]C[/math]. then the slope of g is [math]\frac{BC}{AC}[/math].[br]Step 3 construct a line at [math]B[/math] perpendicular to g and call the intersection point of that line with h [math]D[/math].[br]Then the slope of h is [math]\frac{DE}{AE}[/math][br]Step 5 construct [math]F[/math] such that [math]F[/math] lies on line [math]BC[/math] and [math]\angle DFB=90^\circ[/math][br]Step 6 [math]\angle CAB=\angle FBD[/math] because [math]\angle FBD+\angle DBA+\angle ABC=180^{\circ}\Rightarrow\angle FBD+90^{\circ}+\angle ABC=180^\circ\Rightarrow\angle FBD+\angle ABC=90^\circ[/math] and [math]\angle CAB+\angle ABC+\angle BCA=180^{\circ}\Longrightarrow\angle CAB+\angle ABC+90^\circ=180^{\circ}=\angle CAB+\angle ABC=90^{\circ}[/math] so [math]\angle FBD+\angle ABC=\angle CAB+\angle ABC\Longrightarrow\angle FBD=\angle CAB[/math][br]Step 7 [math]FBD\sim CAB[/math] by AAA(angle-angle-angle)[br]Step 8 [math]AB=DB[/math] because [math]ABD[/math] is a right isosceles triangle[br]Step 9 this implies that [math]CAB[/math] and [math]FBD[/math] are congruent by ASA (angle-side-angle)[br]Step 10 since [math]DF[/math] and [math]AC[/math] are parallel and [math]DE[/math] and [math]CF[/math] are parallel [math]DFCE[/math] is a rectangle so [math]BC=FD=CE[/math]. [math]AE=AC-BC[/math] Additionally since [math]CBF[/math] is a straight line [math]DE=BC+AC[/math].[br]putting it together we get what is shown on the applet[br][br]You can generalize this for arbitrary angles as well[br][br]if you have any questions or comments you can ask me next friday