The resul proved below is probably overkill, but it can be used to solve the problem.
In $\triangle XYZ$, let $M$ be the midpoint of $YZ$. for simplicity, let $x$, $y$, and $z$ be the lengths of the sides opposite $X$, $Y$, and $Z$ respectively. Let $m$ be the length of the median from $X$. Let $\theta=\angle XMY$, and let $\phi=\angle XMZ$. Note that $\theta+\phi=180^\circ$, and therefore $\cos\phi=-\cos\theta$.
By the Cosine Law applied to $\triangle XMY$, we have
$$z^2=(x/2)^2+m^2-2(x/2)m\cos\theta.$$
Similarly,
$$y^2=(x/2)^2+m^2-2(x/2)m\cos\phi.$$
Add, using the fact that $\cos\phi=-\cos\theta$. We get
$$y^2+z^2=\frac{x^2}{2} +2m^2.$$
So we can get an expression for $m^2$ in terms of $x^2$, $y^2$, and $z^2$.
In particular, for your problem, if the two sides and median of the "small" triangle are proportional to the two sides and median of the "big" triangle, then the remaining two sides are proportional, with the same proportionality constant.
The theorem about the length of the median that was proved above can be proved more "geometrically."
Triangles $ABH$ and $ABK$ are right triangles whose hypothenuse is $AB$. So they have the same circumcircle, whose diameter is $AB$. As a consequence, considering that $M$ is the midpoint of the diameter $AB$, we get that $MH=MK$ because they are radii of the same circumference, and then $MHK$ is isosceles.
Without using circles: we can show that, in any right triangle, the median drawn to the hypotenuse is equal to half the hypotenuse. To show it for the right triangle $AHB$,
let us draw a line $ME$ starting from the midpoint $M$, parallel to the
leg $AH$, and ending to the intersection point $E$ with the other leg $HB$. We know that the angle $AHB$ is right. The angles $MEB$ and
$AHB$ are congruent, since they are corresponding angles if we consider the parallel lines $AH$ and $ME$, and the transverse line $HB$. So, the angle $MEB$ is a right angle.
Because the line $ME$ starts from the midpoint $M$ and is parallel to $AH$, it divides the leg $HB$ in two congruent segments $HE$ and $EB$ with equal length. These considerations show that the triangles $MEH$ and $MEB$ are right triangles, have congruent legs $HE$ and $EB$, and a common leg $ME$. This means that these triangles are congruent. We then get that the segments $MH$ and $MB$ are also congruent, since they are corresponding sides (hypothenuses) of these triangles. So we have shown that, in the right triangle $AHB$, the median $MH$ is equal to half the hypotenuse $AB$.
Applying the same procedure to the right triangle $BKA$, we get that $MK$ is congruent with $MA$, so that it also equals half of the hypothenuse $AB$.
We then conclude that $MH=MK$, and so the triangle $MHK$ is isosceles.
Best Answer
Whether you add angles DAC to BAD or ACD, the result ( sum total) is same, equal to 90 degrees. So they are equal.
$$ \angle BAD = \angle ACD $$
similarly
$$ \angle DBA= \angle DAC $$
You said
$$ \angle ADB = \angle CDA = 90^{0} $$
Done.