MHB Necessity of Hypotenuse-Leg Theorem

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There's a theorem in Euclidean Geometry that says: "Let $\Delta$ and $\Delta'$ be two right triangles. If the hypotenuse and a leg of $\Delta$ has the same measure as the hypotenuse and a leg of $\Delta'$, then $\Delta\cong\Delta'$." Wikipedia says this is only a sufficient condition, by I don't see why it wouldn't be necessary as well. If $\Delta\cong\Delta'$, the by $SSS$ criterion, the two hypotenuses are congruent and a side of each. Is this wrong?
 
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Adam1729 said:
There's a theorem in Euclidean Geometry that says: "Let $\Delta$ and $\Delta'$ be two right triangles. If the hypotenuse and a leg of $\Delta$ has the same measure as the hypotenuse and a leg of $\Delta'$, then $\Delta\cong\Delta'$." Wikipedia says this is only a sufficient condition, by I don't see why it wouldn't be necessary as well. If $\Delta\cong\Delta'$, the by $SSS$ criterion, the two hypotenuses are congruent and a side of each. Is this wrong?
The reason that the sufficiency is stated as a theorem is that it is a special case of two triangles in which two sides and a non-included angle are the same for both triangles. This is sometimes referred to as an $A{S}S$ situation, and it does not in general imply congruence. But in this special case, where the non-included angle is a right angle, it is sufficient for congruence.

As for the necessity of the condition, if two triangles are congruent then all the angles and sides of one triangle must be the same as the angles and sides of the other one. So any such condition is always necessary for congruence.
 
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