Question

In an acute angled triangle ABC, AD is the median then prove that AD² = AB²/ 2 + AC²/ 2 - BC²/ 4

Answer 1

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.