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# Find the relation between the longest diagonal and the triangle plane

Question:
Find the angle between the plane of a triangle HAC and the longest diagonal FD in a cube. 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.

If you are interested in watching the video for the solution of this problem, please click the video talk on how to find the relation between the longest diagonal and the triangle plane in a cube.