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On a Problem in Elementary Geometry

Published online by Cambridge University Press:  03 November 2016

F G. Maunsell  and
D. Pedoe
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Extract

The rectangle ABCD is supposed to lie in a vertical plane, with AB horizontal. P is any point in the horizontal plane through CD. Prove that ∠APB < ∠CPD.

Provided that the perpendicular from P on to CD meets CD between C and D, this theorem is true and easily proved by comparing the angles into which a plane through P perpendicular to the rectangle divides the two angles considered. The sine of each of the angles into which ∠APB is divided is less than the sine of the corresponding angle in ∠CPD.

Type
Research Article
Information
The Mathematical Gazette , Volume 25 , Issue 265 , July 1941 , pp. 136 - 140
Copyright
Copyright © Mathematical Association 1941 

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