TITLE: Triangle in perspective to a given triangle but similar to another
QUESTION [7 upvotes]: 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."

REPLY [4 votes]: 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. $\square$