A is right

Because DE is \(\frac{1}{2}\) of CE and FE is \(\frac{1}{2}\) of AE (i.e., corresponding sides have proportional lengths), and angle DEF is shared between /the two triangles, you can see that triangles ACE and FDE must be similar. Similar triangles have the further property that corresponding angles are of equal measure. Thus, for example, angle DFE equals angle CAE, and so FD is parallel to AC.

B is right

Looking at the two smaller triangles in the right half of the figure, you can see that triangle CD F and triangle DEF have collinear and equal “bases” (CD and DE), and share their third vertex (F), which is some fixed distance away from

CE. Because CE is comprised of bases CD and DE of triangles CD F and DEF, respectively, the two triangles have the same height. Because CDF and DEF have equal bases and the same height, they must have the same area. (For a similar reason, triangles ACF and FCE must have equal areas; more on that later.)

Choice E is right

Because triangles ACF and FCE must have equal areas as indicated above, you can see that the area of triangle ACE must be twice that of triangle ACF. Note that FD is parallel to AC due to (true) choice (A), and BF is perpendicular toFD. Therefore, BF must be perpendicular to AC as well. Put differently, AC can be regarded as the base, and BF the height, of triangle ACF. The area of triangle ACF equals ^ times AC x BF. The area of triangle ACE, which is twice that of ACF, must therefore equal AC x BF.

_________________

Get the 2 FREE GREPrepclub Tests