ABCD is a unit square. If CD is moved parallel to AB, and away from AB, continuously, how do you prove that the limit AC/BD is 1?

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Answer 1 · 2018-07-22T06:52:52

See proof.

As DC moves away from A, in the direction #A rarr B#,

#angle C rarr 0#.

At any position, BC = AD = csc C.

#AC^2 = AB^2 + BC^2 - 2 (AB)(BC).cos (180^o- C)#

= 1 +csc^2 C +2 csc C cos C#

Likewise,

#BD^2 = 1 +csc^2 C +2 csc C cos C#. .

#AC^2/(BD^2)#

# = ( 1 +csc^2 C +2 csc C cos C ) /( 1 +csc^2 C +2 csc C cos C#

# = ( sin^2 C + 1 + sin 2C ) / ( sin^2 C + 1 - sin 2C )#

# rarr 1# as # C rarr 0#.

Note that this limit is 1, for the general case of # AB ne CD# abd

the space in between is h. The proof is similar.