Solve it if you can #2

According to a theorem: BD/CD=sin⁡(BAD) ̂ /sin⁡(DAC) ̂ ×AB/AC we replace AB=6 , AC=3 in the formula (we knew that angles BAD and DAC were equal so sin BAD=sin DAC) First=> 2CD=BD then we use cosines theorem: AC^2 = AB^2 + BC^2 -2 AB BC Cos ABC by replacing values of AB and BC and Cos ABC and solving the outcome quadratic equation, we will understand that: Second=> BC=3 3^(1/2) by the First and Second equations we can see that BD=2 (3^(1/2)). and its obvious that distance between D and side AB is equal to BD (1\2) means the distance is = 3^(1\2)

$Drop~DP\perp ~BC. ~What~ we~ need ~is~\color{#D61F06}{ PD}.\\ A~ 30-60-90 ~rt\angle d~ \Delta~is~shown~in~the~sketch.\\ \text{From this we can see that ABC, ABD, APD are all 30-60-90 } \Delta s\\ \therefore~in~\Delta~ACD,~~AD=2\sqrt3.\\ \therefore~in~\Delta~APD,~~\color{#D61F06}{ PD}=2\sqrt3*\dfrac 1 2=1.73$

According to Sin theory, we have: $\frac{|AC|}{sin(B)}=\frac{|AB|}{sin(C)} \rightarrow C=90^{0}$ $BC=\sqrt{|AB|^{2}-|AC|^{2}}$ $d_{D|AB}=\frac{[ABC]}{|AB|+|AC|}=\boxed{1.73}$