Upon the sides AB, AC of a triangle ABC are described equilateral trianglesABD and ACE with their vertices D and E outside the triangle ABC . Prove thatBE = DC .

Answer :

We have ABD and ACE are same equilateral triangle , So

AB =  BD  = DA  = AC  =  CE  = EA 
And from given information we get our figure , As  :

So,
AB =  AC
Then
Triangle ABC is a isosceles triangle , So from base angle theorem  We get
$\angle$ ABC  = $\angle$ ACB 
And
$\angle$ ABD  =  $\angle$  ACE  = 60$°$  ( As these angles from equilateral triangles ABD and ACE )
And
$\angle$ CBD  =  $\angle$ ABC + $\angle$ ABD  ----------------- ( 1 )

And
$\angle$ BCE =  $\angle$ ACB  +  $\angle$ ACE
So,
$\angle$ BCE  =  $\angle$ ABC + $\angle$ ABD  ----------- (  2 )   ( As we know $\angle$ ABC  = $\angle$ ACB  And $\angle$ ABD  =  $\angle$  ACE  )
From equation 1 and 2 , we get
$\angle$ CBD  =  $\angle$  BCE  ---------- ( 3 )

Now in $∆$ CBD and $∆$ BCE

BC  =  BC                                                                               ( Common side )
$\angle$ CBD  =  $\angle$  BCE                                                               ( From equation 3 )
And
BD  = CE                                                                                ( Given )
Hence
$∆$ CBD $\cong$$∆$ BCE                                                                   ( By SAS  rule )
So,

BE  = DC                                                                                 (  By CPCT  )  ( Hence proved )