We have a quadrilateral $ABCD$. Angle $B$ and angle $D$ are right angles. $AB = 10M$. Angle $BAC$ is equal to angle $BCA$, thus triangle $ABC$ is an isosceles triangle. Angle $ACD$ is $30^{\circ}$. We need to find the length of $AD$, which is denoted by $x$.

1. Problem Description

We have a quadrilateral $ABCD$. Angle $B$ and angle $D$ are right angles. $AB = 10M$. Angle $BAC$ is equal to angle $BCA$, thus triangle $ABC$ is an isosceles triangle. Angle $ACD$ is $30^{\circ}$. We need to find the length of $AD$, which is denoted by $x$.

2. Solution Steps

First, since triangle $ABC$ is an isosceles triangle with $BA = BC$, we know that angle $BAC =$ angle $BCA$.

Since the sum of the angles in a triangle is $180^{\circ}$, and angle $ABC = 90^{\circ}$,

angle $BAC$ + angle $BCA$ + angle $ABC = 180^{\circ}$.

So angle $BAC$ + angle $BCA = 180^{\circ} - 90^{\circ} = 90^{\circ}$.

Since angle $BAC =$ angle $BCA$, we have $2 \times$ angle $BCA = 90^{\circ}$, so angle $BCA = 45^{\circ}$.

Since triangle $ABC$ is an isosceles right triangle with $AB = BC = 10M$, we can find $AC$ using the Pythagorean theorem:

$AC^2 = AB^2 + BC^2$

$AC^2 = (10M)^2 + (10M)^2 = 100M^2 + 100M^2 = 200M^2$

$AC = \sqrt{200M^2} = 10\sqrt{2}M$.

In triangle $ACD$, we have angle $ADC = 90^{\circ}$, angle $ACD = 30^{\circ}$. Therefore, angle $CAD = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}$.

We can use trigonometric ratios in the right triangle $ACD$ to find the length of $AD$.

$\cos(30^{\circ}) = \frac{CD}{AC}$. Also, $\sin(30^{\circ}) = \frac{AD}{AC}$.

Since $\sin(30^{\circ}) = \frac{1}{2}$,

$\frac{AD}{AC} = \frac{1}{2}$

$AD = \frac{1}{2} AC$

$AD = \frac{1}{2} (10\sqrt{2}M) = 5\sqrt{2}M$.

So $x = 5\sqrt{2}M$.

3. Final Answer

$x = 5\sqrt{2}M$