Question

ABC is a triangle in which B and C are acute angle. if BE is perpendicular to AC and CF is perpendicular to AB, prove that

Bc^2=ABxBF+ACxCE

Answer 1

Here is your solution

Given :-

ABC is a triangle in which B and C are acute angle. if BE is perpendicular to AC and CF is perpendicular to AB,

To prove :-

BC^2 = AB.BF + AC.CE

Proof :-

In triangle ABC, angle B is acute and CF is perpendicular AB.

AC^2 = AB2 + BC^2 - 2 AB.BF ............... (1)

In triangle ABC, angle B is acute and BE is perpendicular AC.

AB^2 = BC^2 + AC^2 - 2 AC.CE ............. (2)

Adding the above two equations,

=>AC^2 + AB^2 = AB^2 + BC^2 - 2 AB.BF + BC^2 + AC^2 - 2 AC.CE

=>2BC^2 - 2(AB.BF + AC.CE) = 0

=>BC^2 = AB.BF + AC.CE proved

Hope it helps you