In an acute angled triangle ABC, AD is the median then prove that AD² = AB²/ 2 + AC²/ 2 - BC²/ 4
Answers
Step-by-step explanation:
Given
An acute triangle ABC with AD as median
To prove
AD² = AB²/ 2 + AC²/ 2 - BC²/ 4
Construction
Draw AE perpendicular to BC.
Proof
In triangle ABE, by Pythagoras Theorem
AB² = BE² + AE² ........(1)
In triangle ACE, by Pythagoras Theorem
AC² = CE² + AE²..........(2)
In triangle AED, by Pythagoras Theorem
AD² = AE² + ED²..........(3)
Also in triangle ABC
EC - BE
= (ED + DC) - (BD - ED)
= ED + BD - BD + ED ( Since AD is median BD = DC)
= 2ED
or EC - BE = 2ED ,........(4)
Adding (1) and (2) we get
AB² +AC² = BE² + EC² + 2AE²
or AE² = (AB² +AC² - BE² - EC²)/2
Putting value of AE² in (3) we get
AD² = (AB² +AC² - BE² - EC²)/2 + ED²
or AD² = AB²/2 + AC²/2 -BE²/2 - EC²/2 + (EC-BE)²/4 (Using equation (4) )
or AD² = AB²/2 + AC²/2 -BE²/2 - EC²/2 + (BE² + EC - 2EC.BE)/4
or AD² = AB²/2 + AC²/2 -(1/4)*(2BE² + 2EC² - BE² - EC + 2EC.BE)
or AD² = AB²/2 + AC²/2 -(1/4)*(BE² + EC² + 2EC.BE)
or AD² = AB²/2 + AC²/2 -(1/4)*(BE + EC)²
or AD² = AB²/2 + AC²/2 -(1/4)*(BC)²
or AD² = AB²/2 + AC²/2 -(BC)²/4
Hence Proved.