1800-212-7858 (Toll Free)
9:00am - 8:00pm IST all days

or

Thanks, You will receive a call shortly.
Customer Support

You are very important to us

022-62211530

Mon to Sat - 11 AM to 8 PM

# 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 September 2013, 5:39 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

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 Expert 24th September 2013, 6:00 PM
• 1
• 2
• 3
• 4
• 5
• 6
• 7
• 8
• 9
• 10

You have rated this answer /10

RELATED STUDY RESOURCES :

### Latest Questions

ICSE X Geography
Asked by negiankita2003 21st January 2019, 11:38 PM
ICSE VI Physics
Asked by khatoonamina2810 21st January 2019, 11:25 PM
ICSE VI Physics
Asked by khatoonamina2810 21st January 2019, 11:24 PM