In an equilateral triangle $\Delta ABC$ with side $a$, an interior point $D$ is chosen and joined with vertices to form line segments $AD$,$BD$ and $CD$.
Also we know that $\angle ADC=x$, $\angle BDC =y$ and $\angle ADB=z$
Now if a new triangle with line segments $AD$, $BD$ and $CD$ are formed, Is it possible to find the angles of this new triangle?
My try:
Let $AD=p$, $BD=q$ and $CD=r$
By cosine rule we have:
$$\cos z=\frac{p^2+q^2-a^2}{2pq}$$
$$\cos x=\frac{p^2+r^2-a^2}{2pr}$$
$$\cos y=\frac{r^2+q^2-a^2}{2rq}$$
which are three equations in three unknowns $p,q,r$.
Can we solve these?
Let $T$ be the triangle with sides $AD$, $BD$ and $CD$.
Let $R$ denote a rotation by $\frac{\pi}{3}$ about $B$. Let $D'=R(D)$. Therefore, $\triangle BDD'$ is equilateral. Also, $C=R(A)$. Therefore, $CD'=R(A)R(D)=AD$, as rotation preserves length. Therefore, $\triangle CDD'\cong T$. As rotation preserves angles too, $\angle BCD'=\angle BR(A)R(D)=\angle BAD$. Therefore, $\angle DCD'=\angle DCB+\angle BAD= 2\pi-\theta_A-\theta_C-\frac{\pi}{3}=\theta_B-\frac{\pi}{3}$. Therefore, angle opposite to $BD$ in $T = \theta_B-\frac{\pi}{3}$.
Therefore, by symmetry, the angles of $T$ are $\theta_A-\frac{\pi}{3},\ \theta_B-\frac{\pi}{3}$ and $\theta_C-\frac{\pi}{3}$.
$\blacksquare$