Triangle in perspective to a given triangle but similar to another Is it always possible to construct a triangle that is in perspective to a given triangle and have it also be similar to a different given triangle? If you create a triangle in perspective to another, aren't you bound by the angles given by the connecting lines?
By "in perspective" in the sense of the Desargues Theorem: "Corresponding sides of the triangles, when extended, meet at points on a line called the axis of perspectivity. The lines which run through corresponding vertices on the triangles meet at a point called the center of perspectivity."

 A: First, note that a given triangle and a given point of perspectivity determine three rays making some set of angles; we'll say that the "outer" rays form respective angles $\alpha$ and $\beta$ with the "middle" ray. (Having taken opposite rays if necessary, we may assume $\alpha + \beta \leq 180^\circ$.) Also, if a target triangle has a "similar copy" that's perspective with the given triangle, then a "congruent copy" can be put into perspective, as well. (Simply use the center of perspectivity as a center of dilation, and scale the "similar copy" as needed to make it a "congruent copy".) Consequently, we can formulate the task at hand as follows (where we have no need to mention the "original" triangle):

Given angles $\alpha$ and $\beta$ (such that $\alpha + \beta \leq 180^\circ$) and an arbitrary ("target") triangle $\triangle ABC$, find a point $P$ such that $\angle APC = \alpha$ and $\angle BPC = \beta$. 

Easy:
 
The basic construction flow is:

$$\overline{AB} \quad\to\quad \text{isosceles } \triangle AOB \quad\to\quad \bigcirc O \quad\to\quad C^\prime \quad\to\quad \overleftrightarrow{C^\prime C} \quad\to\quad P $$

In more detail:


*

*Construct an isosceles triangle $\triangle AOB$ with $\angle AOB = 2(\alpha+\beta)$ (where we measure the vertex angle "the long way around" if it needs to be greater than $180^\circ$).

*Construct $\bigcirc O$ through $A$ and $B$. By the Inscribed Angle Theorem, all ---and only!--- points on one of the arcs $\stackrel{\frown}{AB}$ make an angle of $\alpha+\beta$ with $A$ and $B$. (Points on the other arc make the supplementary angle.) So, $P$ must be somewhere on this arc.

*Locate $C^\prime$ on the circle, such that $\angle AOC^\prime = 2\alpha$ and $\angle BOC^\prime = 2\beta$.

*Let $P$ be the point where $\overleftrightarrow{C^\prime C}$ meets appropriate arc of the circle. (If $C$ and $C^\prime$ happen to coincide, then $P$ can be any point on that arc.)

*Necessarily, by the Inscribed Angle Theorem, $\angle APC = \alpha$ and $\angle BPC = \beta$, as desired. $\square$

A: Yes, it is possible:
Problem. Given two triangles $ABC$ and $A^{\prime}B^{\prime}C^{\prime}$
and a point $P$ in the plane such that $P$ does not lie on any of the lines
$BC,CA,AB$. Then, there exist three points $A^{\prime\prime},B^{\prime\prime
},C^{\prime\prime}$ on the lines $AP,BP,CP$, respectively, such that triangle
$A^{\prime\prime}B^{\prime\prime}C^{\prime\prime}$ is nondegenerate and
directly similar to triangle $A^{\prime}B^{\prime}C^{\prime}$.
(The word "directly" means that the triangles have the same orientation.)
Solution. Use directed angles modulo $180^{\circ}$. The locus of all points
$Q$ satisfying $\measuredangle B^{\prime}QC^{\prime}=\measuredangle BPC$ is a
circle through $B$ and $C$; denote this circle by $\alpha$. The locus of all
points $Q$ satisfying $\measuredangle C^{\prime}QA^{\prime}=\measuredangle
CPA$ is a circle through $C$ and $A$; denote this circle by $\beta$. The two
circles $\alpha$ and $\beta$ intersect at the point $C^{\prime}$ and another
point (which can still be $C^{\prime}$ if they touch each other); let this
other point be $Q$. Thus, $\measuredangle B^{\prime}QC^{\prime}=\measuredangle
BPC$ (since $Q\in\alpha$) and $\measuredangle C^{\prime}QA^{\prime
}=\measuredangle CPA$ (since $Q\in\beta$) and $\measuredangle A^{\prime
}QB^{\prime}=-\underbrace{\measuredangle B^{\prime}QC^{\prime}}%
_{=\measuredangle BPC}-\underbrace{\measuredangle C^{\prime}QA^{\prime}%
}_{=\measuredangle CPA}=-\measuredangle BPC-\measuredangle CPA=\measuredangle
APB$.
Now, it is not hard to find three points $A^{\prime\prime},B^{\prime\prime
},C^{\prime\prime}$ on the lines $AP,BP,CP$, respectively, such that the
quadrilateral $A^{\prime\prime}B^{\prime\prime}C^{\prime\prime}P$ is directly
similar to the quadrilateral $A^{\prime}B^{\prime}C^{\prime}Q$. Indeed, we
must have either $Q\neq A^{\prime}$ or $Q\neq B^{\prime}$ (or both). We WLOG
assume that $Q\neq A^{\prime}$, since otherwise we can exchange the roles of
$A^{\prime}$ and $B^{\prime}$. Now, pick an arbitrary point $A^{\prime\prime
}\neq P$ on the line $AP$; then construct a point $B^{\prime\prime}$ on the
line $BP$ such that triangle $A^{\prime\prime}PB^{\prime\prime}$ is directly
similar to triangle $A^{\prime}QB^{\prime}$ (this is possible since
$\measuredangle APB=\measuredangle A^{\prime}QB^{\prime}$), and then construct
a point $C^{\prime\prime}$ on the line $CP$ such that triangle $A^{\prime
\prime}PC^{\prime\prime}$ is directly similar to triangle $A^{\prime
}QC^{\prime}$ (this is possible since $\measuredangle CPA=\measuredangle
C^{\prime}PA^{\prime}$). As a result, the quadrilateral $A^{\prime\prime
}B^{\prime\prime}C^{\prime\prime}P$ is directly similar to the quadrilateral
$A^{\prime}B^{\prime}C^{\prime}Q$ (since the former is built of the triangles
$A^{\prime\prime}PB^{\prime\prime}$ and $A^{\prime\prime}PC^{\prime\prime}$,
while the latter is built of the triangles $A^{\prime}QB^{\prime}$ and
$A^{\prime}QC^{\prime}$, which are directly similar to the former two). Hence,
the triangle $A^{\prime\prime}B^{\prime\prime}C^{\prime\prime}$ is directly
similar to the triangle $A^{\prime}B^{\prime}C^{\prime}$. Furthermore, our
triangle $A^{\prime\prime}B^{\prime\prime}C^{\prime\prime}$ is nondegenerate,
because the only way it could be degenerate (while being similar to
$A^{\prime}B^{\prime}C^{\prime}$) would be if it was a single point (i.e., if
we had $A^{\prime\prime}=B^{\prime\prime}=C^{\prime\prime}$), but this is
precluded by $A^{\prime\prime}\neq P$ (indeed, if we had $A^{\prime\prime
}=B^{\prime\prime}=C^{\prime\prime}$, then $A^{\prime\prime}$ would have to
lie on the lines $BP$ and $CP$, and thus equal $P$). Hence, the problem is solved.
