Question

The perpendicular ad on the base bc of triangle abc intersects bc at d so that db=3cd. prove that 2ab^2=2ac^2+bc^2

Answer 1

Draw the rough figure on your copy so that you will understand what am I writing 
In triangle ABD 
AB^2=AD^2+BD^2 
but we have BD = 3/4BC DC=1/4BC 
&AD^2=AC^2-DC^2=AC^2-(BC/4)^2 
therefore AB^2=AC^2-(BC/4)+ (3BC/4)^2 
AB^2=AC^2+BC^2/2 
2AB^2=2AC^2+BC^2

Answer 2

Answer:

See the attachment. Hope it helps uh!