Question

In ΔABC, D is the midpoint of BC and AE⊥BC. If AC>AB, show that `AB^2= AD^2+1/4 BC^2 −BC.DE ` 

Answer 1

In right-angled triangle AED, applying Pythagoras theorem, we have: 

`AB^2=AE^2+ED^2` ...........(1) 

In right-angled triangle AED, applying Pythagoras theorem, we have: 

 

`AD^2=AE^2+ED^2` 

`⇒ AE^2=AD^2-ED^2` ...............(2) 

Therefore, 

`AB^2=AD^2-ED^2+EB^2`   (from(1) and (2)) 

`AB^2=AD^2-ED^2+(BD-DE)^2` 

`=AD^2-ED^2+(1/2BC-DE)^2` 

`=AD^2-DE^2+1/4BC^2+DE^2-BC.DE` 

`=AD^2+1/4BC^2-BC.DE`  

This completes the proof.