Question

prove that 4(AMsq +CNsq) =5 ACsq

Answer 1

Given:-AN=BN=AB/2
BM=CM=BC/2

To prove :-4{(AM)^2+(CN)^2}=5(AC)^2

Proof:- In triangle NBC,Angle B=90°
(NB)^2+(BC)^2=(NC)^2
(AB^2/4)+(BC)^2=(CN)^2 - 1

In triangle ABM, angle B=90°
(AB)^2+(BM)^2=(AN)^2
(AB)^2+(BC^2/4)=(AN)^2 - 2

Adding 1&2
(CN)^2+(AN)^2=AB^2/4+AB^2+BC^2/4+BC^2
CN^2+AN^2=AB^2+4AB^2+BC^2+4BC^2/4
4(CN^2+AN^2)=5AB^2+5BC^2
4(CN^2+AN^2)=5(AB^2+BC^2)
4(CN^2+AN^2)=5AC^2 (since,AB^2+BC^2=AC^2)