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?

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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?

  • 1
    $\begingroup$ 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. $\endgroup$ – Ross Millikan Jan 25 at 15:57
  • $\begingroup$ But $x+y+z=360$ right from the diagram, how can $x,y,z$ be angles of new triangle? $\endgroup$ – 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 $\theta_C-\frac{\pi}{3}$.



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