Why a b = c d ⟹ a + c b + d = a b = c d ?In geometry class, this property was quickly introduced: a b = c d ⟹ a + c b + d = a b = c d It usually avoids some boring quadratic equations, so it's useful but I don't really get it. Seems trivial but I can't prove.How to demonstrate it?

Question

Why       a    b    =      c    d      ⟹              a      +      c              b      +      d        =      a    b    =      c    d  ?In geometry class, this property was quickly introduced:            a      b        =            c      d          ⟹                      a        +        c                    b        +        d              =            a      b        =            c      d      It usually avoids some boring quadratic equations, so it&#039;s useful but I don&#039;t really get it. Seems trivial but I can&#039;t prove.How to demonstrate it?

slapadabassyc · Accepted Answer

a    b    =      c    d    ⇒            a      +      c              b      +      d        =            a      +                        a          d                b                    b      +      d        =            a      b      +      a      d              b      (      b      +      d      )        =      a    b    =      c    d

Caylee Villegas · Accepted Answer

Assume that none of the unknowns is zero. Then, let            a      b        =            c      d        =  kThis implies,  a  =  k  b  ,    c  =  k  d  ∴      a    +    c    =  k  (  b  +  d  )    ⟹                      a        +        c                    b        +        d              =  k

In geometry class, this property was quickly introduced: a/b=c/d=>(a+c)/(b+d)=a/b=c/d It usually avoids some boring quadratic equations, so it's useful but I don't really get it. Seems trivial but I can't prove. How to demonstrate it?

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