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Here is a tricky compass and straightedge construction problem.
Given triangle $\triangle ABC$ and point $D$ on segment $\overline{AB}$, construct point $P$ on line $\overleftrightarrow{CD}$ such that $\angle APB = \angle BAC$.
This configuration appears frequently in Olympiad geometry problems and the diagram is impossible to draw precisely unless you know how to do the construction.
I misread the question the first time I answered this, so let me try again.
The condition that you gave $\angle APB = \angle BAC$ is the degenerate case of the "two angles that inscribe the same arc are equal" theorem. More specifically, if you were to draw the circumcircle of $\triangle APB$, you would see that the tangent to $A$ is $AC$. Also, we know that $$\angle APB = 180^\circ - \angle AID = \angle AIC$$ where $I$ is the intersection of the circumcircle of $APB$ with $BC$. This implies that $$\triangle ABC \sim \triangle IAC.$$ The construction of $I$ isn't that difficult (just copy $\angle B$ onto side $AC$), and everything can be reversed to find $P$.
Construct isosceles triangle QAB with QA equals QB and angle QAB equals 90- BAC/2. The circumcircle of QAB is the locus of points X ( above the line AB) such that angle AXB equals BAC.. The desired point P is the intersection of this circle with CD.
There are 2 possible cases – $P$ can be outside or inside of $\triangle ABC$. Maybe that is why the OP claims that this is trick problem.
Case-1 ($P$ is outside, easier to start with for my $ABC$)
1) Draw line $g$, the perpendicular bisector of $AB$.
2) Draw line $h$, through $A$ and perpendicular to $AC$, cutting $g$ at $O$.
3) Draw circle $k$ using $O$ as center and $OA$ as radius.
4) Let circle $k$ cuts $CD$ (extended) at $P$.
Proof: By angles in alternate segment, $\alpha = \beta$.
Case-2 ($P$ is inside triangle ABC.)
[Continuing from the above]
1) Locate $P’$, the mirror refection of $P$ about $AB$. It should be clear that $\theta = \alpha$.
2) Draw circle $m$, passing through $A, B, P’$, cutting $CD$ at $P”$.
Proof: By angles in the same segment, $\phi = \theta$.
Result follows.
Construct line $AK$ with $K$ on $BC$ (produced, if necessary) and $\angle BAK = \angle ACB$. The circumcircle of $\triangle ABK$ intersects $CD$ (again, produced if needed) at two points $P$ and $P1$. Without loss of generality, let $P$ be on the same side of $AB$ as $K$. Then, $\angle APB = \angle AKB$ and by construction $\angle AKB = \angle BAC$. (since$\angle ABC + \angle BCA + \angle BAC = \angle ABK + \angle BAK + \angle AKB) $.
Thus we are done
