Let ABCD be a tetrahedron with ∠BAD = 60, ∠BAC = 40, ∠ABD = 80, ∠ABC = 70. Prove that the lines AB and CD are perpendicular.
The given angles fix triangles ABC and ABD. Imagine the shared side of this triangle (AB) as a hinge. No angles of the other two triangles are given, so this hinge is free to move.
Swing the hinge to make the triangles coplanar with C inside triangle ABD. It took a surprising amount to trig to show angle ADC=30 but then the sides AB and CD are perpendicular.
Now reflect C over AB to get C'. Segments CC' and AB are perpendicular and ACBC' is a kite, so D, C, and C' are collinear.
Now looking straight down at triangle ABD, let C rotate about AB. The image of C projected down on the plane of ABD lies along CC' so it is proved that AB and CD are orthogonal.
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Posted by Jer
on 2021-02-03 08:45:20 |
Solution
