Question

$\huge \sf { \fcolorbox{yellow}{g}{†Q†}}$
in this figure , BN and CM are medians of a ∆ABC right-angled at A. Prove that 4(BN²+CM²)= 5BC²


no spamz❌​

Answer 1

Answer:

values changed see to it ... and squares mention..

Step-by-step explanation:

BL is median

⇒ AL=CL=

2

1

AC →(1)

CM is median

⇒ AM = MB = 21

AB →(2)

InΔBAC

(BC) 2 =(AB) 2 +(AC) 2

InΔBAC

(BL) 2 =(AB) 2 +( 2AC ) 2

4BL 2 =4AB 2 +(AC) 2

In ΔMAC,(CM) 2 =(AM) 2 +(AC) 2

(CM) 2 =( 2AB ) 2+(AC) 2

4CM 2=(AB) 2 +(AC) 2

NOW,(BC) 2 =(AB) 2 +(AC) 2 →(1)

4BC 2 =4(AB) 2+(AC) 2 →(2)

4CM 2=AB 2+4AC 2 →(3)

ADD (2)&(3)

4BC 2 +4CM 2=5AB 2 +5AC 2

4(BL 2+CM 2 )=5BC 2

♥

Answer 2

Step-by-step explanation:

Hope my attachment Is clear... ✌

happy to help :)

doubt irundha kelu... mistakes irundhaalum solu (: