In triangle ABC angle A =60. Prove that BC sq=AB sq+ACsq - AB.AC

1.	In triangle ABC angle A =60. Prove that BC sq=AB sq+ACsq - AB.AC
Answer» Given: In {tex}\\triangle{/tex}ABC,To prove: BC2\xa0= AB2\xa0+ AC2\xa0- AB.ACConstruction: Draw CE\xa0{tex}\\perp{/tex} ABProof: In {tex}\\triangle{/tex}BECBC2\xa0= CE2\xa0+ BE2\xa0..(i)In {tex}\\triangle{/tex}ACE, AC2\xa0= CE2\xa0+ AE2CE2\xa0= AC2\xa0- AE2\xa0..(ii)from (i) and (ii)BC2\xa0= AC2\xa0- AE2\xa0+ (AB - AE)2\xa0({tex}\\because{/tex} BE = AB - AE)BC2\xa0= AC2\xa0+ AB2\xa0- 2AB.AE - 2AB{tex}\\frac{1}{2}{/tex}ACSince side opposite to 30° angel is half hypoteneuse.In {tex}\\triangle{/tex}ACE, AE = {tex}\\frac{1}{2}{/tex}ACBC2 = AB2 + AC2 - AB.AC

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