Let $\triangle {ABC}$ cu $m(\angle C)=90^{\circ}$ and $D\in [BC], E\in [AC]$ s.t. $\frac{BD}{AC}=\frac{AE}{CD}=k$.
If $BE\cap AD=\{O\}$ show that $m(\angle BOD)=60^{\circ}$ iff $k=\sqrt{3}$.
I tried to prove it with trigonometry, with tangent, but there are a lot of computations. Also I constructed perpendiculars but also I have to do a lot of computations and I am stuck.
Let us prove that $k=\sqrt 3\implies \angle BOD=\angle AOE=60^\circ.$
First, construct lines perpendicular to lines $AC$ and $BC$ at points $E$ and $D$. Their intersection is point $F$. Also construct line $OG$ perpendicular to $AC$ at point $G$ and line $OH$ perpendicular to line $BC$ at point $H$.
To simplify calculations, introduce the following lengths: $AC=b$, $EF=CD=x$, $OG=p$ and $OH=q$. Our first task is to calculate $p$ and $q$ in terms of $b$ and $x$.
It is given that:
$$BD=AC\sqrt{3}=b\sqrt3\tag{1}$$
$$AE=CD\sqrt{3}=EF\sqrt{3}=x\sqrt{3}\tag{2}$$
It is obvious that $\triangle ACD\sim\triangle AGO$ and therefore:
$$\frac{GO}{CD}=\frac{AG}{AC}=\frac{AC-GC}{AC}$$
$$\frac{p}{x}=\frac{b-q}{b}\tag{3}$$
It is also obvious that $\triangle ECB\sim\triangle OHB$ and therefore:
$$\frac{EC}{OH}=\frac{BC}{BH}$$
$$\frac{AC-AE}{OH}=\frac{BD+CD}{BD+CD-CH}$$
$$\frac{b-x\sqrt 3}{q}=\frac{b \sqrt 3+x}{b \sqrt 3+x-p}\tag{4}$$
Now solve (3) and (4) for $p$ nad $q$ and you get:
$$p=\frac{x^2}{b^2+x^2}(b\sqrt 3 + x) \tag{5}$$
$$q=\frac{b^2}{b^2+x^2}(b-x\sqrt 3)\tag{6}$$
Let us now compare red angles $\alpha_1=\angle EAO$ and $\alpha_2=\angle EFO$:
$$\tan\alpha_1=\frac{CD}{AC}=\frac xb\tag{7}$$
$$\tan\alpha_2=\frac{EG}{HD}=\frac{AC-AE-CG}{CD-CH}=\frac{b-x\sqrt 3-q}{x-p}\tag{8}$$
If you replace (5) and (6) into (8) you get:
$$\tan\alpha_2=\frac{b-x\sqrt 3-\frac{b^2}{b^2+x^2}(b-x\sqrt 3)}{x-\frac{x^2}{b^2+x^2}(b\sqrt 3 + x)}=\frac{x}{b}$$
In other words, red angles $\alpha_1$ and $\alpha_2$ are equal! The rest is simple: because of this, quadrilateral $AEOF$ is cyclic (angles $\alpha_1$ and $\alpha_2$ above the same quadrilateral side are equal). Consequentially, green angles $\angle AOE$ and $\angle AFE$ are also equal:
$$\angle AOE = \angle AFE$$
But $\tan\angle AFE=\sqrt 3\implies \angle AFE=60^\circ$. It means that $\angle AOE=\angle BOD=60^\circ$.
I'll leave the detailed proof of the opposite statement to you.
HINT: You can show in exactly the same way that quadrilateral $AEOF$ is conclyclic, just replace $\sqrt 3$ with $k$ in previous expressions. It means that angles $\angle EFA$ and $\angle EOA$ are equal to $60^\circ$. Unknown factor $k$ is simply equal to $\tan\angle EFA=\tan 60^\circ=\sqrt 3$