MHB Necessity of Hypotenuse-Leg Theorem

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The Hypotenuse-Leg Theorem states that if two right triangles have congruent hypotenuses and one leg, then the triangles are congruent. While this is a sufficient condition for congruence, it is not necessary, as congruence requires all sides and angles to be equal. The discussion highlights that the theorem is a special case of the Angle-Side-Side (ASS) situation, which does not generally imply congruence. However, in right triangles, the theorem holds true due to the right angle being included. Therefore, while the condition is sufficient for congruence, it is not necessary in all cases.
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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.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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