190. If any four numbers be proportional, and if from the first two, _a_ and _b_, any two homogeneous expressions of the same degree be formed; and if from the last two, two other expressions be formed, in precisely the same manner, the four results will be proportional. For example, if _a_: _b_ ∷ _c_: _d_, and if 2_aaa_ + 3_aab_ and _bbb_ + _abb_ be chosen, which are both homogeneous with respect to _a_ and _b_, and both of the third degree; and if the corresponding expressions 2_ccc_ + 3_ccd_ and _ddd_ + _cdd_ be formed, which are made from _c_ and _d_ precisely in the same manner as the two former ones from _a_ and _b_, then will 2_aaa_ + 3_aab_ : _bbb_ + _abb_ ∷ 2_ccc_ + 3_ccd_ : _ddd_ + _cdd_ _a_ To prove this, let --- be called _x_. _b_ _a_ _a_ _c_ Then, since --- = _x_, and --- = ---, _b_ _b_ _d_ _c_ it follows that --- = _x_. _d_ But since _a_ divided by _b_ gives _x_, _x_ multiplied by _b_ will give _a_, or _a_ = _bx_. For a similar reason, _c_ = _dx_. Put _bx_ and _dx_ instead of _a_ and _c_ in the four expressions just given, recollecting that when quantities are multiplied together, the result is the same in whatever order the multiplications are made; that, for example, _bxbxbx_ is the same as _bbbxxx_. Hence, 2_aaa_ + 3_aab_ = 2_bxbxbx_ + 3_bxbxb_ = 2_bbbxxx_ + 3_bbbxx_ which is _bbb_ multiplied by 2_xxx_ + 3_xx_ or _bbb_ (2_xxx_ + 3_xx_)[29] Similarly, 2_ccc_ + 3_ccd_ = _ddd_ (2_xxx_ + 3_xx_) Also, _bbb_ + _abb_ = _bbb_ + _bxbb_ = _bbb_ multiplied by 1 + _x_ or _bbb_(1 + _x_) Similarly, _ddd_ + _cdd_ = _ddd_ (1 + _x_) Now, _bbb_ : _bbb_ ∷ _ddd_ : _ddd_ [29] If _bx_ be substituted for _a_ in any expression which is homogeneous with respect to _a_ and _b_, the pupil may easily see that _b_ must occur in every term as often as there are units in the degree of the expression: thus, _aa_ + _ab_ becomes _bxbx_ + _bxb_ or _bb_(_xx_ + _x_); _aaa_ + _bbb_ becomes _bxbxbx_ + _bbb_ or _bbb_(_xxx_ + 1); and so on. Whence (186), _bbb_(2_xxx_ + 3_xx_): _bbb_(1 + _x_) ∷ _ddd_(2_xxx_ + 3_xx_): _ddd_(1 + _x_), which, when instead of these expressions their equals just found are substituted, becomes 2_aaa_ + 3_aab_: _bbb_ + _abb_ ∷ 2_ccc_ + 3_ccd_: _ddd_ + _cdd_. The same reasoning may be applied to any other case, and the pupil may in this way prove the following theorems: If _a_ : _b_ ∷ _c_ : _d_ 2_a_ + 3_b_ : _b_ ∷ 2_c_ + 3_d_ : _d_ _aa_ + _bb_ : _aa_ - _bb_ ∷ _cc_ + _dd_ : _cc_ - _dd_ _mab_ : 2_aa_ + _bb_ ∷ _mcd_ : 2_cc_ + _dd_