Through a point on the [hypotenuse](https://artofproblemsolving.com/wiki/index.php/Hypotenuse) of a [right triangle](https://artofproblemsolving.com/wiki/index.php/Right_triangle), lines are drawn [parallel](https://artofproblemsolving.com/wiki/index.php/Parallel) to the legs of the triangle so that the triangle is divided into a [square](https://artofproblemsolving.com/wiki/index.php/Square) and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The [ratio](https://artofproblemsolving.com/wiki/index.php/Ratio) of the area of the other small right triangle to the area of the square is 
$\textbf {(A)}\ \frac{1}{2m+1} \qquad \textbf {(B)}\ m \qquad \textbf {(C)}\ 1-m \qquad \textbf {(D)}\ \frac{1}{4m} \qquad \textbf {(E)}\ \frac{1}{8m^2}$
Solution 1

WLOG, let a side of the square be $1$. Simple angle chasing shows that the two right triangles are [similar](https://artofproblemsolving.com/wiki/index.php/Similar_triangles). Thus the ratio of the sides of the triangles are the same. Since $A = \frac{1}{2}bh = \frac{h}{2}$, the base of the triangle with area $m$ is $2m$. Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is the height of the other triangle. $x = \frac{1}{2m}$, and the area of that triangle is $\frac{1}{2} \cdot 1 \cdot \frac{1}{2m} = \frac{1}{4m}\ \text{\boxed{D}}$.

Solution 2 (Video Solution)
[https://youtu.be/HTHveknJFpk](https://artofproblemsolving.comhttps://youtu.be/HTHveknJFpk)
[https://m.youtube.com/watch?v=TUQsHeJ6RSA&feature=youtu.be](https://artofproblemsolving.comhttps://m.youtube.com/watch?v=TUQsHeJ6RSA&feature=youtu.be)

Solution 3

From the diagram from the previous solution, we have $a$, $b$ as the legs and $c$ as the side length of the square. WLOG, let the area of triangle $A$
be $m$ times the area of square $C$.
Since triangle $A$ is similar to the large triangle, it has $h_A = a(\frac{c}{b}) = \frac{ac}{b}$, $b_A = c$ and \[[A] = \frac{bh}{2} = \frac{ac^2}{2b} = m[C] = mc^2\]
Thus $\frac{a}{2b} = m$
Now since triangle $B$ is similar to the large triangle, it has $h_B = c$, $b_B = b\frac{c}{a} = \frac{bc}{a}$ and \[[B] = \frac{bh}{2} = \frac{bc^2}{2a} = nc^2 = n[C]\]
Thus $n = \frac{b}{2a} = \frac{1}{4(\frac{a}{2b})} = \frac{1}{4m}$. $\text{\boxed{D}}$.
~ Nafer

Solution 4 (process of elimination)
Simply testing specific triangles is sufficient.
A triangle with legs of 1 and 2 gives a square of area $S=\frac{2}{3}\times\frac{2}{3}=\frac{4}{9}$. The larger sub-triangle has area $T_1=\frac{\frac{2}{3}\times\frac{4}{3}}{2}=\frac{4}{9}$, and the smaller triangle has area $T_2=\frac{\frac{2}{3}\times\frac{1}{3}}{2}=\frac{1}{9}$. Computing ratios you get $\frac{T_1}{S}=1$ and $\frac{T_2}{S}=\frac{1}{4}$. Plugging $m=1$ in shows that the only possible answer is $\text{\boxed{D}}$
~ Snacc

Solution 5

WLOG, let the length of the square be $1$ (Like Solution 1). Then the length of the larger triangle is $2m$. Let the length of the smaller triangle be $x$.
Therefore, since $A = B$ (try to prove that yourself), $1 = 2mx$ or $x = 1/2m$
The area of the other triangle is $1/4m$.
From here, the answer is $\text{\boxed{D}}$.