Final answer:
Two right triangles are congruent if one leg and the altitude to the hypotenuse of one triangle are congruent to a leg and the altitude to the hypotenuse of the other triangle. This can be shown by superimposing the triangles, aligning the congruent leg and altitude, revealing that the remaining parts align as well, proving congruence by leg-leg.
Step-by-step explanation:
To prove that two right triangles are congruent if a leg and the altitude to the hypotenuse of one triangle are congruent to a leg and the altitude to the hypotenuse of the other triangle, we can use the principle of superposition.
Let's consider two right triangles, Triangle 1 (ABC) and Triangle 2 (DEF), where:
- AC and DF are congruent legs (AC ≅ DF).
- BD and EG are the altitudes to hypotenuses (BD ≅ EG).
Since BD and EG are altitudes to the hypotenuses, we know that they create two additional right triangles within our original triangles, namely AB and BC for Triangle 1, and DE and EF for Triangle 2. By the definition of congruent triangles, these sets of smaller triangles must also be congruent, respectively. Similarly, since the legs AC and DF are congruent, their respective smaller triangles, namely ABC and DEF, are congruent as well which implies the hypotenuses BC and EF are congruent.
We can see that the two triangles share a common leg and altitude, thus they can be superimposed on top of each other aligning these two parts. After this superposition, the remaining parts will also align, therefore proving the triangles are congruent by the rules of superposition and right triangle congruence (leg-leg).