SSA ProofDate: 03/14/2001 at 14:54:05 From: Brent Gillett Subject: Re: Geometry SSA congruence How do I prove the hypotenuse-leg congruence for triangles? My professor suggested using the S.S.A. theorem Date: 03/14/2001 at 17:00:41 From: Doctor Rob Subject: Re: Geometry SSA congruence Thanks for writing to Ask Dr. Math, Brent. If two triangles have one angle, one adjacent side, and the opposite side equal, then they may or may not be congruent. That means that using S.S.A. is tricky. Suppose, for example, you were given acute <A and sides b and a. Then the picture might look like this: C _,-' _,o' _,-' . b _,-' . _,-' d. _,-' . _,-' . A o-----------------------o----------- X If you know some trigonometry, you'll realize that d = b*sin(<A). Now draw a circle with radius a and center C. Point B must lie on line AX and on the circle. There are several cases: Case 1: a < d. The circle and line don't intersect, and there are no triangles with these parameters. Case 2: a = d. The circle and line are tangent at X, so B = X, and there is just one triangle with these parameters. In this case, <ABC is a right angle. Case 3: d < a < b. The circle and line intersect in two points B and B', with B' between A and X, and B to the right of X. Now there are two triangles with these parameters, ABC and AB'C, and you can't prove congruence. Case 4: b <= a. The circle and the line intersect in two points B and B', but B' is at or to the left of A, so the angle drawn above is an exterior angle to that triangle. Thus there is just one triangle with these parameters, and you can prove congruence. If <A is a right angle, and so d = b, the following cases hold: Case 5: a <= b. No triangle. Case 6: b < a. There is just one triangle, and you can prove congruence. If <A is obtuse, then following cases hold: Case 7: a <= b. No triangle. Case 8: b < a. There is just one triangle, and you can prove congruence. Thus the theorem should say, THEOREM (S.S.A.): If two triangles ABC and DEF have BC congruent to EF, AC congruent to DF, and <A congruent to <D, then: 1. if AC >= BC, the triangles are congruent. 2. if BC = AC*sin(<A), then the triangles are congruent. 3. if AC*sin(<A) < BC < AC, then the triangles may or may not be congruent. In your case, you know that the hypotenuse AC is longer than either leg, including BC, so Part 1 of the above theorem applies. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ |
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