Question

.........................

Answer 1

Answer:

$2$

Step-by-step explanation:

Let $[n]$ denote the area of some polygon $n$.

The sum of the area of quadrilaterals BCDE and MNDE is equal to $[ABC]-([MBE]+[NDC]+AMN])$ (big triangle subtracting the three smaller triangles that aren't part of BCDE or MNDE).

Let the area of triangle ABC be $\alpha$.

Triangle AMN is similar to triangle ABC. Because their side lengths are in ratio 1:2 respectively, the ratio of the area of triangle AMN to triangle ABC is $1:2^2=1:4$.

Therefore, $\displaystyle [AMN]=\frac{1}{4}[ABC]=\frac{1}{4}\alpha$.

Now for triangle NDC, NDC is similar to triangle AMC. Since M is the midpoint of AB, AMC is half the area of ABC. The ratio of the side lengths between triangle NDC and triangle AMC is $1:2$ because the D is the midpoint of MC, so it follows the ratio of their areas must be $1:4$, respectively.

Therefore, $\displaystyle[NDC]=\frac{1}{4}[AMC]=\frac{1}{8}[ABC]=\frac{1}{8}\alpha$.

We can use the same concept for triangle MEB. Triangle MEB is similar to triangle ANB, and the ratio of their sides is also $1:2$, since E is the midpoint of BN. Their areas are in the ratio $1:2^2=1:4$ and we have $\displaystyle [MBE]=\frac{1}{4}[ANB]=\frac{1}{8}\alpha$.

Thus,

$\displaystyle \frac{[ABC]}{[BCDE]+[MNDE]}=\frac{\alpha}{\alpha-\frac{1}{4}\alpha-\frac{1}{8}\alpha-\frac{1}{8}\alpha}=\frac{\alpha}{0.5\alpha}=\frac{1}{\frac{1}{2}}=\boxed{2}$