In a triangle ABC the angles at B and C are acute. If BE and CF be drawn perpendiculars on AC and AB respectively, prove that BCxBC= ABxBF+ACxCE.
Asked by | 24th Sep, 2013, 05:39: PM
In triangle ABC, angle B is acute and CF is perpendicular AB.
So, AC2 = AB2 + BC2 - 2 AB.BF
In triangle ABC, angle B is acute and BE is perpendicular AC.
So, AB2 = BC2 + AC2 - 2 AC.CE
Adding the above two equations,
AC2 + AB2 = AB2 + BC2 - 2 AB.BF + BC2 + AC2 - 2 AC.CE
2BC2 - 2(AB.BF + AC.CE) = 0
BC2 = AB.BF + AC.CE
Answered by | 24th Sep, 2013, 06:00: PM
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