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 BC2= AB x BF + AC x CE.
In Δ BCE by Pythagoras theorem, we have
⇒ BC2 = BE2 + CE2 ... (1)
Again, in Δ ABE by Pythagoras theorem
AB2 = BE2 + AE2
⇒ BD2 = AB2 - AE2
On putting value of BE2 in (1), we get
BC2 = AB2 - AE2 + CE2
⇒ BC2 = AB2 - (AC - CE)2 + CE2
⇒ BC2 = AB2 - AC2 - CE2 + 2AC. CE + CE2
⇒BC2 = AB2 - AC2 + 2AC. CE -----------(2)
In Δ BCF by Pythagoras theorem, we have
⇒ BC2 = BF2 + CF2 ... (3)
Again, in Δ ACF by Pythagoras theorem
AC2 = CF2 + AF2
⇒ CF2 = AC2 - AF2
On putting value of CF2 in (3), we get
BC2 = AC2 - AF2 + BF2
⇒ BC2 = AC2 - (AB - BF)2 + BF2
⇒ BC2 = AC2 - AB2 - BF2 + 2AB. BF + BF2
⇒BC2 = AC2 - AB2 + 2AB. BF----------(4)
Adding (2) and (4)
2BC2 = 2AB. BF+2AC.CE
⇒BC2 = AB. BF+AC.CE
Hence Proved