I recently encountered 'construction problems' in geometry. These were quite new to me and I didn't know the requirements they expected and prerequisites to solve them. I'll explain with an example.
Let $A, B, C, D$ be four given points on a line $l$. Construct a square such that two of its parallel sides or their extensions go through $A$ and $B$ and the other two sides (or their extensions) go through $C$ and $D$.
Construct a line $CH$ passing through $C$ such that $\angle BCG = \tan^{-1}\left(\frac{a}{b}\right)$. Construct a line $DF$ parallel to $CH$ passing through $D$. Construct a line $BF$ perpendicular to $CH$ and passing through $B$. Construct a line $AH$ through $A$ parallel to $BF$.
The intersection of these four lines forms the required square i.e. $KHGF$ is a square.
Now, to prove that it indeed is a square, construct $BI$ perpendicular to $AH$ through $B$ and $CJ$ perpendicular to $DF$ through $C$.
Let $CD = b$, $AB = a$
Let $\tan^{-1}\left(\frac{a}{b}\right) = \theta$
$$\angle JDC = \angle GCB = \theta$$
$$FG = JC = CD\cdot\sin\theta = b\sin\theta$$
Similarly,
$$\angle IBA = \angle GCB = \theta$$
$$IB = AB\cdot\cos\theta = a\cos\theta$$
$$\frac{JC}{IB} = \frac{b\sin\theta}{a\cos\theta} = \frac{b}{a}\tan\theta = \frac{b}{a}\cdot \frac{a}{b} = 1$$
$$\implies IB = JC = FG = GH$$
Also, since $CH||DK$, and $BF||AK$, opposite sides are parallel and all angles are right angles. Therefore, $KFGH$ is a square.
Turns out, this quite simple solution is not correct. There is some elaborate procedure of constructions described in my book. I just want to know, what is expected in these sorts of questions and how to approach them.