Question

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 BC2= AB x BF + AC x CE.

Answer 1

here is your correct me

Answer 2

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