Question

AD is a median of a triangle ABC and AM is perpendicular to BC. prove that AC²=AD²+BC*DM+1/4BC².

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)