 # Right side hypotenuse - why is a right angle needed?

From Wikipedia:

RHS (Right-angle-Hypotenuse-Side), also known as HL (Hypotenuse-Leg): If two right-angled triangles have their hypotenuses equal in length, and a pair of shorter sides are equal in length, then the triangles are congruent.

I can’t picture why a right angle is needed. To me:

Given two triangles with congruent hypotenuses. If the angle opposite this hypotenuse is congruent then so are the triangles. The angle shouldn’t have to be a right angle.

What am I missing?

Please ignore. Site won’t let me edit or delete.

Whoops.

Can’t, sorry. Just following up to agree that you are essentially correct in identifying this right-triangle congruence rule as basically just a special case of the more general “Side-Angle-Side” triangle congruence rule.

That is, any two triangles which are equal in one angle, and also equal in each of the two sides adjacent to that angle, are congruent. This is true whether the angle is right or oblique.

But the hypotenuse isn’t adjacent to the right angle.

IIUC, what the OP is talking about is “Side-Side-Angle,” which isn’t sufficient to establish congruence in general, but it does work for right triangles.

What you actually need for SSA to be enough is for the side adjacent to the given angle to be smaller than the side opposite the angle. This certainly works when the angle is right or obtuse, since that must be the largest angle, and hence opposite the largest side. It sometimes works for acute angles, too. But it’s usually easier to prove congruence using one of the other postulates than to show which side is larger.

I’ve always thought of this as a special case of SSS. Using Pythagoras if the hypothenuse and a leg are congruent then the third sides have to be congruent.

Please provide for me a proof that two triangles which are not right triangles can be proven to be congruent using SSA.

@ Those pleading for the SSA “theorem” — Do you find any flaw in the proof that all triangles are equilateral ?

Take a triangle with side B as the base and draw a ray (with some angle) coming from the right vertex ®.

Adjust a compass to the length of side A, place it on the left vertex (L), and draw a circle. If A>B (Chronos’ constraint), there can be only be one point of intersection, since R is inside the circle and the ray originates from there. The point of intersection is the third vertex and is unambiguous.

If A<=B, then the ray originates on or outside the circle. Thus there must be 0 or 2 intersections. If there are 0, then we never had a triangle. If there are 2, then there is ambiguity in the third vertex and we can construct two non-congruent but SSA triangles.

Left off a small but important bit.

If the ray is outside the circle, then we could have 1 intersection as well. This only happens when the ray is tangent to the circle, and thus a right angle to side A (the radius of the circle). So SSA works in the particular case of a right triangle.

:o Did nobody click the link and diagnose the flaw? It’s a nice puzzle!

Maybe they didn’t get a chance, between all of the other things they’re doing. But really, plus or minus one unread webpage, what difference does it make? Ya’ll are writing the case wrong. As we learned in junior high, it’s not hard to remember that there is no A-S-S theorem.