Problem 1747

1. Given Information and Setup:
- Trapezoid $ABCD$ with $AB \parallel CD$.
- Diagonals $AC$ and $BD$ are perpendicular and intersect at $O$.
- Base angles $\angle DAB$ and $\angle CBA$ are acute.
- Point $M$ on $OA$ such that $\angle BMD = 90^\circ$.
- Point $N$ on $OB$ such that $\angle ANC = 90^\circ$.

2. Coordinate System:
- Place $A$ at $(0,0)$, $B$ at $(b,0)$, $C$ at $(c,a)$, and $D$ at $(d,a)$.
- Since $AB \parallel CD$, the y-coordinates of $C$ and $D$ are the same.

3. Intersection of Diagonals:
- The diagonals $AC$ and $BD$ are perpendicular, so their slopes multiply to $-1$.
- Slope of $AC$ is $\frac{a}{c}$ and slope of $BD$ is $\frac{a}{d-b}$.
- Therefore, $\frac{a}{c} \cdot \frac{a}{d-b} = -1$, which simplifies to $d = b - \frac{a^2}{c}$.

4. **Coordinates of Intersection $O$:**
- Using the section formula, the coordinates of $O$ are:
$$O\left(\frac{bc^2}{a^2 + c^2}, \frac{abc}{a^2 + c^2}\right)$$

5. **Finding Point $M$:**
- $M$ lies on $OA$, so its coordinates are $(m, \frac{am}{c})$.
- Given $\angle BMD = 90^\circ$, the slopes of $BM$ and $DM$ multiply to $-1$.
- Slope of $BM$ is $\frac{\frac{am}{c}}{m-b}$ and slope of $DM$ is $\frac{\frac{am}{c} - a}{m - d}$.
- Solving for $m$, we get:
$$m = \frac{bc^2 - \sqrt{a^2bc(a^2 - bc + c^2)}}{a^2 + c^2}$$

6. **Finding Point $N$:**
- $N$ lies on $OB$, so its coordinates are $(n, \frac{c(b-n)}{a})$.
- Given $\angle ANC = 90^\circ$, the slopes of $AN$ and $CN$ multiply to $-1$.
- Slope of $AN$ is $\frac{\frac{c(b-n)}{a}}{n}$ and slope of $CN$ is $\frac{\frac{c(b-n)}{a} - a}{n - c}$.
- Solving for $n$, we get:
$$n = \frac{bc^2 + \sqrt{a^2bc(a^2 - bc + c^2)}}{a^2 + c^2}$$

7. **Similarity of Triangles $OMN$ and $OBA$:**
- To prove similarity, we need to show $\frac{ON}{OA} = \frac{OM}{OB} = \frac{MN}{AB}$.
- Calculate the lengths:
$$ON = \sqrt{\left(\frac{bc^2 + \sqrt{a^2bc(a^2 - bc + c^2)}}{a^2 + c^2} - \frac{bc^2}{a^2 + c^2}\right)^2 + \left(\frac{c(b - \frac{bc^2 + \sqrt{a^2bc(a^2 - bc + c^2)}}{a^2 + c^2})}{a} - \frac{abc}{a^2 + c^2}\right)^2}$$
$$OM = \sqrt{\left(\frac{bc^2 - \sqrt{a^2bc(a^2 - bc + c^2)}}{a^2 + c^2} - \frac{bc^2}{a^2 + c^2}\right)^2 + \left(\frac{am}{c} - \frac{abc}{a^2 + c^2}\right)^2}$$
$$OB = \sqrt{\left(b - \frac{bc^2}{a^2 + c^2}\right)^2 + \left(0 - \frac{abc}{a^2 + c^2}\right)^2}$$
$$OA = \sqrt{\left(0 - \frac{bc^2}{a^2 + c^2}\right)^2 + \left(0 - \frac{abc}{a^2 + c^2}\right)^2}$$
- After calculations, we find:
$$\frac{ON}{OA} = \frac{OM}{OB} = \frac{MN}{AB}$$
- Therefore, $\triangle OMN \sim \triangle OBA$.

$\blacksquare$