Question

AD is the median of a ΔABC, prove that AB + BC + CA > 2AD​

Answer 1

PLEASE MARK IT AS THE BRAINLIEST ANSWER

Answer 2

Answer:

you have to use the theorum : sum of two sides is greater than the third

Step-by-step explanation:

in Δ ACD:  AC + CD > AD =  AC + 1/2 BC > AD  ----------- Eq 1

then in ΔADB :  AB + BD > AD = AB + 1/2 BC > AD ----------------- Eq 2

now Eq1 + Eq2 :

( AC + 1/2 BC > AD ) + ( AB + 1/2 BC > AD )

= AB + AC + ( 2 X 1/2 BC ) > 2 X AD

= AB + AC + BC > 2AD

hence proved ....

pls mark brainliest pls