The angle bisector

I haven't learn the bisector theorem yet so I have another solution without using it. Because AD is the bisector of angle BAC, which is a right angle, so $\widehat{BAD} = \widehat{CAD} = 45^{\circ}$ . Immediately, I think about isosceles right triangle. Thus I draw $DE\perp AB$ and $DF\perp AC$ . By SA for right triangles I got $\triangle AED \cong \triangle AFD$ , so DE = DF. But DE and DF also the altitude of $\triangle ADB$ and $\triangle ADC$ , and I can find the area of $\triangle ADB$ , $\triangle ADC$ and $\triangle ABC$ (let $[\triangle ABC]$ be the area of triangle ABC): $[\triangle ABD] = \frac{AB \cdot DE}{2} = \frac{3DE}{2} \\ [\triangle ACD] = \frac{AC \cdot DF}{2} = 3DE\\ [\triangle ABC] = \frac{AB \cdot AC}{2} = 9 \\ [\triangle ABD] + [\triangle ACD] = [\triangle ABC]$ Hence: $\frac{3DE}{2} + 3DE = 9 \\ \Rightarrow DE = 2$ Proving that ADE is an isosceles right triangle is easy and thanks to Pythagorean theorem, we are done: $AD = \sqrt{DE^{2} + DE^{2}} = 2\sqrt{2} \approx 2.83$ . Also, you can draw altitudes from B and C, then use area of triangles, too (and sorry for my bad English, I am not good at English)

$\text{By the angle bisector theorem, we know that }\frac{BD}{DC}=\frac{1}{2}$ $\text{Therefore, we can let }BD = x \text{ and }DC=2x$ $\text{Then, we have the equation: }3x=\sqrt{45} \Rightarrow x=\sqrt{5}$ $\text{By the cosine rule, }5=9+AD^2-6\cos{45^\circ}AD$ $\text{Solving, AD = 2.83}$

Inverse tan(6/3) = 63.43 degrees = angle B

Angle BAD = 1/2 of Angle BAC = 45 degrees

Angle ADB = 180 -45 - 63.43 = 71.57 degrees

Law of Sines : AD/sin 63.43 = 3/sin 71.57

AD = 2.83