In a triangle ABC angel B=45 .Prove that AC^2=AB^2+BC^2-4AREA(triangle

1.	In a triangle ABC angel B=45 .Prove that AC^2=AB^2+BC^2-4AREA(triangle ABC).
Answer» Construction: Draw AD {tex}\\perp{/tex}\xa0BCProof: In right {tex}\\triangle{/tex}ADC,By using pythagoras theorem, we getAC2 = AD2 + CD2 ...(i)In right {tex}\\triangle{/tex}ADB,By using pythagoras theorem, we getAB2 = AD2 + BD2{tex}\\Rightarrow{/tex}AB2\xa0- BD2 = AD2Putting in (i), we getAC2 = AB2\xa0- BD2 + CD2{tex}\\Rightarrow{/tex}\xa0AC2 = AB2 + CD2\xa0- BD2= AB2+ (CD - BD)(CD + BD) = AB2 + (CD - BD) BC= AB2 + (BC - BD - BD).BC = AB2 + BC2\xa0- 2BC.BD= AB2 + BC2\xa0- 2.AD.BC [ {tex}\\because{/tex}\xa0AD = BD]= AB2 + BC2\xa0- 4ar({tex}\\triangle{/tex}ABC).

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