Can we find angles of the new triangle?

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?

• Your title talks about finding angles while the body seems to assume the angles are known and you want to find $p,q,r$. You can go either direction. – Ross Millikan Jan 25 at 15:57
• But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle? – Umesh shankar Jan 25 at 16:02 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 $$\theta_C-\frac{\pi}{3}$$.
$$\blacksquare$$