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A folded corner
The crease creates equal angles when the corner is folded across it.
ABCD is a rectangular piece of paper. Corner C is folded along the crease BF, where F is on DC. The fold is a wide one: the corner swings right past the edge AB, and C lands at C' outside the paper. Angle C'BF = 58°. Find angle BFC'.
Show the step-by-step solution
- ∠CBF = ∠C'BF = 58° (folded angles are equal)
- ∠BFC = 180° - 90° - 58° = 32° (angle sum of triangle BCF)
- ∠BFC' = ∠BFC = 32° (folded angles are equal)
Ans: 32°