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Find the angle between the plane of a triangle and the longest diagonal in a cube

Question:

EFGH-ABCD is a cube. HA is a diagonal in the plane HDAE. AC is a diagonal in the plane ABCD. CH is a diagonal in the plane CGHD. FD is the longest diagonal in the cube. What is the angle between the plane of the triangle HAC and the line segment FD?

Solution:

Connect ED

Because EF is perpendicular to plane EHDA, and ED is on the plane EHDA, so EF is perpendicular to ED.

So ED is the projection of inclined line segment FD on the plane EHDA.

Because EHDA is a square, so ED is perpendicular to HA. (This is a property of a square.)

Because the projection ED is perpendicular to HA, so the inclined line segment FD is also perpendicular to HA (This is a theorem of three perpendicular lines).

Connect DB

Because FB is perpendiculat plane BCDA, and DB is on the plane BCDA, so FB is perpendicular to DB.

So, DB is the projection of FD on the plane BCDA.

Because BCDA is a square, so DB is perpendicular to AC. (This is a property of a square.)

Because DB is the projection of FD on the plane BCDA, so FD is perpendicular to AC; (This is a theorem of three perpendicular lines).

Because HA and AC are two intersected lines and two intersected lines make a plane, so FD is perpendicular to the plane determined by the two intersected line segments, HA and AC.

So, FD is perpendicular to plane HAC.