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Question Number 12019 by Nayon last updated on 09/Apr/17

Answered by ajfour last updated on 10/Apr/17

Commented by ajfour last updated on 10/Apr/17

AC^( 2) +BD^2 =CF^( 2) +h^2 +ED^2 +h^2 =(CD−CE)^2 +h^2 +(CD−DF)^2 +h^2 = CD^2 +CE^2 +CD^2 +DF^( 2) +h^2 +h^2 −2CD(CE+DF) = (CE^2 +h^2 )+(DF^( 2) +h^2 )+2CD^2 −2CD(CD−AB) = BC^( 2) +AD^2 +2AB.CD

$${AC}^{\:\mathrm{2}} +{BD}^{\mathrm{2}} ={CF}^{\:\mathrm{2}} +{h}^{\mathrm{2}} +{ED}^{\mathrm{2}} +{h}^{\mathrm{2}} \\ $$$$=\left({CD}−{CE}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} +\left({CD}−{DF}\right)^{\mathrm{2}} +{h}^{\mathrm{2}} \\ $$$$=\:{CD}^{\mathrm{2}} +{CE}^{\mathrm{2}} +{CD}^{\mathrm{2}} +{DF}^{\:\mathrm{2}} +{h}^{\mathrm{2}} +{h}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{CD}\left({CE}+{DF}\right) \\ $$$$=\:\left({CE}^{\mathrm{2}} +{h}^{\mathrm{2}} \right)+\left({DF}^{\:\mathrm{2}} +{h}^{\mathrm{2}} \right)+\mathrm{2}{CD}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{CD}\left({CD}−{AB}\right) \\ $$$$=\:{BC}^{\:\mathrm{2}} +{AD}^{\mathrm{2}} +\mathrm{2}{AB}.{CD} \\ $$

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