Is $ABC$ similar with $A'B'C'$, where $A', B', C'$ are the projections of $A, B, C $ on a plane $\pi $.

Question

Let $ABC $ a triangle and $A'$, $B'$, $C'$ its projections on a plane $\pi $ that not contain the triangle $ABC $.

Is $ABC$ similar with $A'B'C'$,i.e their angles are congruent.

Why is

$\mathcal {A}_{\triangle A'B'C'}=\mathcal {A}_{\triangle ABC}\cdot cos (\angle (ABC), \pi) $?

Answer 1

In general, $ABC$ and $A'B'C'$ are not similar. Suppose, for instance, that $\pi$ is orthogonal to the plane containing $A$, $B$, and $C$. Then $A'$, $B'$, and $C'$ are collinear and therefore $A'B'C'$ is a degenerate triangle.