Given that in ΔABC, we have
AD⊥BC and BD=3CD
In right angle triangles ADB and ADC, we have
AB2=AD2+BD2...(i)
AC2=AD2+DC2...(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB2−AC2=BD2−DC2
=9CD2−CD2[∴BD=3CD]
=8CD2
=8(BC4)2[Since,BC=DB+CD=3CD+CD=4CD]
Therefore, AB2−AC2=BC22
⇒2(AB2−AC2)=BC2
⇒2AB2−2AC2=BC2
∴2AB2=2AC2+BC2