Show that $CD\perp AB $. Let $\triangle ABC $ with all angles $<90°$. Let $A_1$ the middle of $  [BC] $. 
If $\angle BAA_1=30°$ and $D\in [AB] $ s.t. $CD=AB $ show that $CD\perp AB $.
My idea : I draw $A_1T\perp AB $, $T\in [AB]$. 
I have to show that $A_1T $ is the middle line in $\triangle CBD $ $\Leftrightarrow$ $A_1T=\frac {CD}{2}=\frac {AB}{2} $.
But $A_1T=\frac{AA_1}{2} $. So I need to prove that $ AA_1=AB $. Now I am stuck.

Let $A = (1,0),$ $B=(0,0),$ and $C = \left(2 - \frac2{\sqrt3},\frac23\right).$
Since $2 - \frac2{\sqrt3} \approx 0.845299,$
angles $\angle CAB$ and $\angle CBA$ are acute,
and since the distance from $C$ to $\left(\frac12,0\right)$ is greater than
$\frac12,$ the angle $\angle ABC$ also is acute.
Then $A_1 = \left(1 - \frac1{\sqrt3},\frac13\right).$
Line $AA_1$ has the equation $y = \frac1{\sqrt3}(1 - x),$
which you can confirm by plugging in the coordinates of $A$ and $A_1,$
and the slope of $AA_1$ is 
$-\frac1{\sqrt3} = \tan(-30^\circ),$
so $\angle BAA_1 = 30^\circ.$
The circle of radius $AB = 1$ around $C$ has equation
$$\left(x + \frac2{\sqrt3} - 2\right)^2 + \left(y - \frac23\right)^2 = 1.$$
Put $y = 0$ in this equation and we find that
$$\left(x + \frac2{\sqrt3} - 2\right)^2 + \frac49 = 1,$$
$$x + \frac2{\sqrt3} - 2 = \pm \sqrt{\frac59},$$
$$x = 2 - \frac2{\sqrt3} \pm \frac{\sqrt5}3.$$
We have $\frac{\sqrt5}3 \approx 0.745356,$
so one of the solutions for $x$ is 
$x_1 = 2 - \frac2{\sqrt3} - \frac{\sqrt5}3 \approx 0.099943.$
Let $D = (x_1, 0).$ Then $D$ lies on the segment $AB$ and $CD = AB = 1,$
but $CD$ is not perpendicular to $AB.$
In fact, if $C$ is any point $(x,y)$ that satisfies $y = \frac1{\sqrt3}(2 - x)$
and $0 < x < 1,$ then all three angles of the triangle are acute and 
$\angle BAA_1 = 30^\circ.$
But the only way we have $D$ on the segment $AB$ with $CD = AB$ and $CD \perp AB$
is if $C = \left(1 - \frac{\sqrt3}2,1\right).$
It seems some necessary condition is missing from your problem statement.
