PathFinder007 wrote:
In the figure above, AB is perpendicular to BC and BD is perpendicular to AC. Which of the following relationships must be true.
VERITAS PREP OFFICIAL SOLUTION:What makes this question particularly difficult is the abstraction it entails; no numbers are provided in either the stimulus or the answer choices. But you do know multiple things about these triangles that should help you make sense of them:
1) Pythagorean Theorem \((a^2 + b^2 = c^2)\), which means that:
\((AB)^2 + (BC)^2 = (AC)^2\)
\((BD)^2 + (DC)^2 = (BC)^2\)
and
\((BD)^2 + (AD)^2 = (AB)^2\)
2) Area of a triangle equals \(\frac{1}{2}(base)(height)\), so since the area could be taken with AB as the height and BC as the base *OR* with BD as the height and AC as the base, you have:
\((AB)(BC) = (BD)(AC)\)
Now, using the Substitution Method for systems of equations you could look to get all the terms down to just AB, BD, and BC as you see in the answer choices, you could also pick numbers that will work with Pythagorean Theorem and check the answer choices. Since this is a right triangle but not necessarily isosceles, try the first set of Pythagorean triplets you can find, which is likely the 3:4:5 triangle. If you say that AB = 3, BC = 4, and AC = 5, you can then also say that BD needs to be 12/5, since the area calculations have to match with (AB)(BC) = (AC)(BD). Then if you plug in for the answer choices, you'll see that squaring (12/5) will lead to a messy fraction while squaring the other terms leads to integers, so none of A, D, and E can be correct. Then try B: \(\frac{1}{(BD)^2}=\frac{1}{(BC)^2}+\frac{1}{(AB)^2}\) would be:
\(\frac{25}{(144)}=\frac{1}{16}+\frac{1}{9}\)
Finding a common denominator on the right, you'll see that B is correct, as:
as:
\(\frac{25}{(144)}=\frac{9}{144}+\frac{16}{144}\).