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?

Question

1941 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-22T06:52:52

See proof.

As DC moves away from A, in the direction $A \to B$ $A rarr B$ ,

$∠ C \to 0$ $angle C rarr 0$ .

At any position, BC = AD = csc C.

$A C^{2} = A B^{2} + B C^{2} - 2 (A B) (B C) . cos (180^{o} - C)$ $AC^2 = AB^2 + BC^2 - 2 (AB)(BC).cos (180^o- C)$

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

Likewise,

$B D^{2} = 1 + {csc}^{2} C + 2 csc C cos C$ $BD^2 = 1 +csc^2 C +2 csc C cos C$ . .

$A \frac{C^{2}}{B D^{2}}$ $AC^2/(BD^2)$

$= \frac{1 + {csc}^{2} C + 2 csc C cos C}{1 + {csc}^{2} C + 2 csc C cos C}$ $= ( 1 +csc^2 C +2 csc C cos C ) /( 1 +csc^2 C +2 csc C cos C$

$= \frac{{sin}^{2} C + 1 + sin 2 C}{{sin}^{2} C + 1 - sin 2 C}$ $= ( sin^2 C + 1 + sin 2C ) / ( sin^2 C + 1 - sin 2C )$

$\to 1$ $rarr 1$ as $C \to 0$ $C rarr 0$ .

Note that this limit is 1, for the general case of $A B \neq C D$ $AB ne CD$ abd

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