$[A'B'C'] = [ABC] \cos \theta$ where $\theta$ is the angle between the plane containing $ABC$ and the projection plane $P.$

Question

Let $\theta$ be the angle between the plane containing $ABC$ and the projection plane $P.$ If $A'B'C'$ is the projection of $ABC$ onto $P,$ then $[A'B'C'] = [ABC]\cos \theta.$

What's the fastest way to prove this? Every time I set up variables for $A, B, C, A', B', C',$ computing the necessary cross products becomes a pain. Specifically, when I take the magnitude of the cross product, I can't get anywhere because I'm taking the magnitude of a sum of vectors that can't be combined. There must be a really quick and painless method, but I'm just not seeing it.

Answer 1

First prove it for the special case when one side of the triangle is parallel to the intersection line of the two planes.
Then the projection of that side has the same length and the corresponding altitude will scale down by a factor of $\cos\theta$ at projection.

Then, for the general case simply cut your triangle along a line parallel to the intersection of the planes.