Question

∆ DEF is an equilateral triangle seg DP | seg EF E-P-F prove that DP2 = 3EP2​

Answer 1

Answer:

SINCE, the DEF is an equilateral triangle

∴ DE = EF = DF  .......eq(1)

NOW,

   ∵ The DP is perpendicular to EF.

   => The DP is bisecting the side EF.

∴ EP=PF

NOW,

       In ΔDEF we have ,

=>  DP2+EP2=DE2       ...(By Pythagoras's theorem)

     =>  DP2+EP2=EF2   .....( By using eq(1) )

     =>  DP2+EP2=   [2EF^2]

      => DP2+EP2=     4EP^2

         

DP2=3EP2

HENCE PROOVED

TQ......////////

       

Step-by-step explanation: