Question

please give answer of this question fast

Q. 18

Answer 1

To prove- AC²+ AB²= 2AD+1/2BC²

Proof-     let the distance between E and D be x, and A and E be y

⇒AD²=x²+y²

AC²= AE²+EC²=y²+[A/2 +x]²___________1

AB²=AE²+EB²=y²+[A/2-x]²_____________2

_2+_1⇒AB²+AC²=2[x²+y²]+ A²/2

⇒AC²+ AB²= 2AD+1/2BC²

Hence Proved

Hope this helped you Bhumika!!!

Answer 2

In right ∆AEC,

AC^2 = AE^2 + CE^2 [By Pythagoras theorem] → (1)

 In right ΔAED,

AD^2 = AE^2 + DE^2 [By Pythagoras theorem] 

⇒ AE^2 = AD^2 – DE^2

 Hence equation (1) becomes, 

AC^2 = [AD^2 – DE^2 ] + CE^2       

  = AD^2 – DE^2 + (DC + DE)^2      

  = AD2 – DE^2 + DC^2 + 2 (DC)(DE) + DE^2       

= AD^2 + DC^2 + 2 (DC)(DE)       

= AD2 + (BC/2)^2 + 2 (BC/2)(DE) 

∴ AC^2 = AD^2 + (BC/2)^2 + (BC)(DE)