Question 275833
<font face="Garamond" size="+2">


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ AC\ =\ \frac{4}{3}AB]


so


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ AB\ =\ \frac{3}{4}AC]


and since


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ BC\ =\ AC\ -\ AB]


we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ BC\ = \frac{1}{4}AC]


Therefore


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ AB\ =\ 3BC]


Furthermore,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ CD\ =\ BD\ -\ BC]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ CD\ =\ 6BC\ -\ BC]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ CD\ =\ 5BC]


Now we have both AB and CD in terms of BC, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{AB}{CD}\ =\ \frac{3BC}{5BC}\ =\ \frac{3}{5}]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
</font>