Find an angle in a triangle given a cevian and a couple of equal segments

Question

I found the following problem in an university entrance exam from the year 1979 How to calculate $x$ if it holds $NA = BC,\ NB = NC$?

I tried to solve it by extendeding $BC$ from $B$ until I get a segment $DC$ of the same lenght as $AC$, but then I can't continue anymore, because I think it lacks more data, but I'm not so sure. Is it possible to find the value of $x$ only with these data?

Answer 1

Let $M$ be the midpoint of $\overline{BC}$ (which is necessarily also the foot of the perpendicular from $N$ to $\overline{BC}$), and let $P$ be the foot of the perpendicular from $N$ to $\overleftrightarrow{AB}$.

Note that $|NP|=|NA|\sin30^\circ=\frac12|NA|=|MC|$. From this, we may conclude $\triangle PNB\cong\triangle MCN$, and thus $\angle PNB=\theta$. Since $\angle ANB=2\theta$, we have $$3\theta=60^\circ \quad\to\quad \theta=20^\circ$$ Done!

... or not ...

As noted in @JohnOmeilan's answer and comment, the alternative solution ($\theta=60^\circ$) arises if $P$ lies between $A$ and $B$:

In this case, all four marked segments in the original image are congruent.

Answer 2

The problem may be solved using just the lines already there. First, set the red line lengths to be $y$ (i.e., $y = |NB| = |NC|$) and the purple line lengths to be $z$ (i.e., $z = |NA| = |BC|$), as shown below

Since $\triangle BCN$ is isosceles, then $\measuredangle BCN = \measuredangle CBN = x$, so $\measuredangle BNC = 180^{\circ} - 2x$. Also,

$$\measuredangle BNA = 2x, \; \; \measuredangle NBA = 180^{\circ} - (30^{\circ} + 2x) = 150^{\circ} - 2x \tag{1}\label{eq1A}$$

Using the law of sines in $\triangle BCN$, plus that $\sin(180^{\circ}-2x) = \sin(2x)$ and $\sin(2x) = 2\sin(x)\cos(x)$ (e.g., see List of trigonometric identities), gives

$$\begin{equation}\begin{aligned} \frac{\sin(x)}{y} & = \frac{\sin(180^{\circ}-2x)}{z} = \frac{\sin(2x)}{z} \\ \frac{\sin(x)}{y} & = \frac{2\sin(x)\cos(x)}{z} \\ \frac{z}{2y} & = \cos(x) \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

Next, using \eqref{eq2A}, the law of sines in $\triangle ABN$, that $\sin(30^{\circ}) = \frac{1}{2}$ and $\sin(\theta + 90^{\circ}) = \cos(\theta)$, we have

$$\begin{equation}\begin{aligned} \frac{\sin(30^{\circ})}{y} & = \frac{\sin(150^{\circ}-2x)}{z} = \frac{\cos(60^{\circ}-2x)}{z} \\ \frac{z}{2y} & = \cos(60^{\circ}-2x) \\ \cos(x) & = \cos(60^{\circ}-2x) \\ \pm x+(360^{\circ})k & = 60^{\circ}-2x,\;\;k\in\mathbb{Z} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$

With $k=0$, we get $2$ solutions for $x$,

$$x = 60^{\circ} - 2x \; \; \to \; \; x = 20^{\circ}, \qquad -x = 60^{\circ} - 2x \; \; \to \; \; x = 60^{\circ} \tag{4}\label{eq4A}$$

All other values of $k$ lead to $x$ values which are out of range (e.g., $k = 1$ gives $x=-100^{\circ}$ or $x=140^{\circ}$).