<s>[INST] prove the angle is 90 degrees Given the trianle ABC,draw AD, where D is the middle of BC.If the angle BAD is 3 times the angle DAC and the angle BDA is 45 degrees,then prove that the angle BAC is 90 degrees.
I tried to draw a parallel line to BA and compare congruent trianges ,after extending AD to meet the parallel line, but i got no result in proving that BAC is 90 degrees
 [/INST] Here is a trigonometric proof.
Let $x=\angle DAC$; then $3x=\angle BAD$. Let $y=BD=DC$. Then some angle chasing gives $\angle ABC=135^{\circ}-3x$ and $\angle ACB=45^{\circ}-x$. By the Law of Sines,
\begin{align*}
  \frac{y}{\sin 3x} &= \frac{AD}{\sin(135^{\circ}-3x)} \\
  \frac{y}{\sin x} &= \frac{AD}{\sin(45^{\circ}-x)},
\end{align*}
so that 
$$\frac{\sin(135^{\circ}-3x)}{\sin 3x} = \frac{\sin(45^{\circ}-x)}{\sin x}.$$
Expanding the numerators and simplifying gives
$$\cot 3x+1 = -1+\cot x-1,\text{ or }\cot 3x = -2+\cot x.$$
Writing $\cot 3x = \frac{\cos 3x}{\sin 3x}$ and using the triple-angle formula, then rewriting in terms of $\cot x$, gives
$$\frac{\cot^3 x-3\cot x}{3\cot^2 x-1} = -2+\cot x.$$
Finally, collecting terms and simplifying, we get
$$2 \cot ^3 x-6 \cot ^2 x+2\cot x +2 = 2(\cot x-1)(\cot^2 x-2\cot x-1)=0.$$
Thus $\cot x=1$ or $\cot x = 1\pm\sqrt{2}$. The only one of these that produces an angle for $\angle BAC$ between $0$ and $180^{\circ}$ is $\cot x = 1+\sqrt{2}$, which gives $x = \frac{\pi}{8}$ so that $\angle BAC = \frac{\pi}{2}$.
</s>