x² + y² - 4x - 8y - 16 = 0
+ 16 + 16
x² + y² - 4x - 8y = 16
x² - 4x + y² - 8y = 16
(x² - 4x + 4) + (y² - 8y + 16) = 16 + 4 + 16
(x² - 2x - 2x + 4) + (y² - 4y - 4y + 16) = 20 + 16
[x(x) - x(2) - 2(x) - 2(-2)] + [y(y) - y(4) - 4(y) - 4(4)] = 36
[x(x - 2) - 2(x - 2)] + [y(y - 4) - 4(y - 4)] = 36
(x - 2)(x - 2) + (y - 4)(y - 4) = 36
(x - 2)² + (y - 4)² = 36
36 36 36
¹/₃₆(x - 2)² + ¹/₃₆(y - 4)² = 1
Center: (2, 4)
You have two angles congruent, plus a side that's NOT between them.
I guess you'd call that situation " AAS " for "angle-angle-side".
That's what you have, and it's NOT enough to prove the triangles
congruent. There can be many many different pairs of triangles
that have AAS = AAS.
So there's no congruence postulate to cover this case, because they're
not necessarily.
