I need help in understanding the problem and figure out a way to have a process when questions like these show up.
Thanks in advance!
Hi,
The faster, perhaps optimal GRE approach
Since CE, the hypotenuse of \triangle{CEF}, corresponds to EF, but CE>EF, \triangle{CDE} is the larger of the two similar triangles. Therefore its area is greater.
The longer approach below proves this.
- Determine the relevant similar triangles and their corresponding sides and angles as shown in the image below.
- The corresponding sides of two similar triangles are proportional. Use the corresponding angles to identify the matching sides and establish a proportion.
\frac{x°_{CEF}}{x°_{CDE}}=\frac{z°_{CEF}}{z°_{CDE}}
Also,
\frac{x°_{CEF}}{x°_{CDE}}=\frac{CF}{DE}
And,
\frac{z°_{CEF}}{z°_{CDE}}=\frac{EF}{CE}
Hence,
\frac{CF}{DE}=\frac{EF}{CE}
And by extension
{CF}=\frac{EF\times{DE}}{CE}
Considering Qa and subbing CF into Qa,
Qa = \frac{CF\times{EF}}{2}~\longrightarrow\frac{EF\times{DE}\times{EF}}{CE\times{2}}
Considering Qb
Qb = \frac{CE\times{DE}}{2}
Manipulate Qa and Qb by dividing both by DE, multiplying both by 2.
Qa =\frac{EF\times\cancel{DE}\times{EF}}{CE\times\cancel{2}}~~~~~~~~~~~~~~~\frac{CE\times\cancel{DE}}{\cancel{2}}=Qb
Finally,
Qa ={EF}^2~~~~~~~~~~~~~~~{CE}^2=Qb
Now, it becomes clear that Qa < Qb since in \triangle{CEF}, EF < CE.
Formally,
\operatorname{Area}(\triangle CDE)
=
\left(\frac{CE}{EF}\right)^2
\times
\operatorname{Area}(\triangle CEF)