1. Zadatak

Neka je $\angle BAN=\alpha$ i $\angle ANM=\varphi$.

Promatrani izraz je

$$\begin{aligned} \frac{\operatorname{tg}\varphi}{\operatorname{tg}\alpha} &=\frac{\sin\varphi}{\cos\varphi}\cdot\frac{\cos\alpha}{\sin\alpha}\\ &=\frac{\sin\varphi}{\sin(90^\circ-\varphi)}\cdot\frac{\sin(90^\circ-\alpha)}{\sin\alpha}\\ &=\frac{\sin\varphi}{\sin(\angle BNM)}\cdot\frac{\sin(\angle ABN)}{\sin\alpha}. \end{aligned}$$

Pritom smo koristili da je trokut $ABN$ pravokutan (Talesov teorem).

Dobiveni izraz možemo pisati u obliku

$$\frac{\sin\varphi}{\sin\alpha}\cdot\frac{\sin(\angle ABN)}{\sin(\angle BNM)}.$$

Primjenom poučka o sinusima na trokut $ANM$ dobivamo

$$\frac{\sin\varphi}{\sin\alpha}=\frac{|AM|}{|MN|},$$

a primjenom na trokut $BMN$

$$\frac{\sin(\angle ABN)}{\sin(\angle BNM)}=\frac{|MN|}{|BM|}.$$

Stoga je

$$\frac{\sin\varphi}{\sin\alpha}\cdot\frac{\sin(\angle ABN)}{\sin(\angle BNM)}=\frac{|AM|}{|MN|}\cdot\frac{|MN|}{|BM|}=\frac{|AM|}{|BM|},$$

a to očito ne ovisi o položaju točke $N$.