Question

in a triangle ABC the angles at B and C are acute if BE and CF are drawn perpendiculars on AC and AB respectively. Prove that BC^2=AB*BF + AC*CE. ​

Answer 1

Answer:

Answer:BC^2=AB.BF+AC.CE

Step-by-step explanation:

In ABC, angle b is acute and CF is parallel to ab

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

in ABC angle b is acute and BE is parallel to AC

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

adding 1 and 2

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

=2BC^2-2(AB.BF+AC.CE)=0

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

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

Step-by-step explanation:

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Answer 2

Answer:Here is your solution

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.CEProof :-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

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