Question

If a, b, c, d are in proportion, prove that: (ma? + nb) : (mc + nd?) = (ma? - nb) : (mc? : (me² - nd²)
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Answer 1

Answer:

If (ma + nb) : b : : (mc + nd) : d, prove that a, b, c, d are in proportion.

Step-by-step explanation:

Consider, (ma+nb):b :: (mc+nd):d

Consider, (ma+nb):b :: (mc+nd):d⟹

Consider, (ma+nb):b :: (mc+nd):d⟹ b

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb)

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb)

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) =

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd)

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd)

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd)

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnd

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnb

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bc

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcb

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba =

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba = d

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba = dc

Consider, (ma+nb):b :: (mc+nd):d⟹ b(ma+nb) = d(mc+nd) d(ma+nb)=b(mc+nd)dma+dnb=bmc+bnddma−bmc=bnd−dnbm(da−bc)=n(bd−db)m(da−bc)=n(0)m(da−bc)=0da−bc=0 ---- As m cannot be equal to 0∴da=bcba = dc