bpolley00 wrote:Can someone in more detail touch on being able to put those two together?
most likely, the notation is at least partly to blame; a lot of people forget that you can type "√" by pressing alt+v (if you have a mac) or, failing that, by just googling "square root sign" and then copying and pasting.
the first statement says that (AC)^2 = 2(AD)^2.
if you take the square root of both sides, you get AC = (√2)AD.
the second statement implies that AC = BC, since those are the only two sides of triangle ABC that can actually be the same. (the hypotenuse of a right triangle can't equal one of the legs.) so, each of AC and BC is equal to (√2)AD.
the only side you're missing now (as far as the sides necessary to find the triangles' areas) is AB.
three ways to go from here:
1/
remember that it's data sufficiency. because triangle ABC is a 45º-45º-90º triangle (isosceles right triangle), you know that it has to have
some constant proportions (even if you don't remember what they are). so, AB is some constant multiple of AC or BC, so it's AD times
something.
if that's the case, then the area of triangle ABD is (base)(height)/2 = (some # times AD)(AD)/2 = some number times (AD)^2.
the area of triangle ABC is (base)(height)/2 = (AD√2)(AD√2)/2 = AD^2. that's either definitely equal or definitely not equal to the area of ABD (it can't be both), so, sufficient.
2/
using the pythagorean theorem, actually find the value of AB:
(AB)^2 = (AD√2)^2 + (AD√2)^2
this gives AB = 2AD.
so, then, the area of triangle ABD is (base)(height)/2 = (2AD)(AD)/2 = (AD)^2.
this is the same as the area of triangle ABC (calculated above), so you get "yes" = sufficient.
3/
if you have the proportions for 45º-45º-90º triangles memorized, you can also find AB that way.
the hypotenuse of such a triangle is √2 times either of the legs, so AB = (√2)(AD√2) = 2AD.
same as #2 from here.
