CBSE Class 10CBSE
Account
Register

Share

Books Shortlist

# Solution - In the given figure, the incircle of ∆ABC touches the sides BC, CA and AB at D, E, F respectively. Prove that AF + BD + CE = AE + CD + BF = 1/2(perimeter of ΔABC) - CBSE Class 10 - Mathematics

ConceptCircles Examples and Solutions

#### Question

In the given figure, the incircle of ∆ABC touches the sides BC, CA and AB at D, E, F respectively. Prove that AF + BD + CE = AE + CD + BF = \frac { 1 }{ 2 } ("perimeter of ∆ABC")

#### Solution

You need to to view the solution
Is there an error in this question or solution?

#### APPEARS IN

NCERT Mathematics Textbook for Class 10
Chapter 10: Circles
Q: 0 | Page no. 0

#### Reference Material

Solution for question: In the given figure, the incircle of ∆ABC touches the sides BC, CA and AB at D, E, F respectively. Prove that AF + BD + CE = AE + CD + BF = 1/2(perimeter of ΔABC) concept: Circles Examples and Solutions. For the course CBSE
S