Question

If in triangle ABC , AD is the median and AM perpendicular to BC, then prove that AB^2 + AC^2=2AD^2 + 1/2 BC^2

Answer 1

In right ΔAMC, AC2 = AM2 + CM2 [By Pythagoras theorem] → (1)
In right ΔAMD, AD2 = AM2 + DM2 [By Pythagoras theorem]
⇒ AM2 = AD2 – DM2
Hence equation (1) becomes,
AC2 = [AD2 – DM2 ] + CM2
          = AD2 – DM2 + (DC + DM)2
        = AD2 – DM2 + DC2 + 2 (DC)(DM) + DM2
        = AD2 + DC2 + 2 (DC)(DM)
       = AD2 + (BC/2)2 + 2 (BC/2)(DM)
∴ AC2 = AD2 + (BC/2)2 + (BC)(DM)

Answer 2

We have AD as the median and AM perpendicular to BC

Since, AD is median, D is the midpoint of BC. hence,

BD = CD

Now, since AM is perpendicular to BC, so AMC is a right angled triangle. Hence,

AC² = AM² + MC²

Now, MC² can be written as (DM + DC)²

=> AC² = AM² + (DM + DC)²

=> AC² = AM² + DM² + DC² + 2DC × DM

(since, (a + b)² = a² + b² + 2ab)

Now, 2DC = BC (since D is the midpoint of BC)

=> AC² = AM² + DM² + DC² + BC × DM

Now, AM² = AD² - DM² (By Pythagoras Theorem, in ∆AMD)

=> AC² = AD² - DM² + DM² + DC² + BC × DM

=> AC² = AD² + BC × DM + DC²

Now, DC = 1/2 BC

=> DC² = (1/2 × BC)²

=> DC² = 1/4 × BC²

So put DC² as 1/4 BC² and we have

AC² = AD² + BC × DM + 1/4 × BC²

Hence Proved :)