Proof 3.3.19
Proof of hypotenuse-leg criterion for congruence
Consider two right triangles ABC and DEF with right angles A and D, AB congruent to DE, and BC congruent to EF.
Construct a point X on the ray starting at C and going through A, such that AX is congruent to DF. Then connect X to B to form right triangle ABX.
Then we can see that AB is congruent to itself, angle CAB is congruent to angle XAB (since they are both right angles), and line XB is congruent to line BC (since XF is congruent to EF and EF is congruent to BC).
But since we know that ABC and ABX are right triangles, we know that angle ABC and angle ABX are congruent, else AB would no longer be perpendicular to XAC and we would no longer have two right triangles. Then by SAS, we can see that triangles ABX and ABC are congruent.
Now since XB is congruent to BC, we know that triangle XBC is isosceles by Definition 20. By Euclid's fifth proposition, we know that the base angles, angle BCX and angle BXC, are congruent. Then we can see that angles X, C, and F are congruent since we defined AX to be congruent to DF.
Then, by replacing X with F, B with E, and A with D in triangle ABX, we can easily see that AB is congruent to DE, angle CAB is congruent to angle FDE, FE is congruent to BC, DF is congruent to AC, angle ABC is congruent to angle DEF, and angle C is congruent to angle F. Therefore, it is clear that two right triangles with a congruent leg and hypotenuse are congruent.