Math, asked by konikasharma518, 8 months ago

Upon the sides AB, AC of a AABC
are described equilateral triangles
ABD and ACE with their vertices D
and E outside the AABC. Prove that
BE = DC.
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Answers

Answered by titan2218
5

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 ,

So,

AB =  AC

Then

Triangle ABC is a isosceles triangle , So from base angle theorem  We get

∠ ABC  = ∠ ACB And

∠ ABD  =  ∠  ACE  = 60°  ( As these angles from equilateral triangles ABD and ACE )

And ∠ CBD  =  ∠ ABC + ∠ ABD  ---------- ( 1 )

And ∠ BCE =  ∠ ACB  +  ∠ ACE

So,

∠ BCE  =  ∠ ABC + ∠ ABD  ----------- (  2 )   ( As we know ∠ ABC  = ∠ ACB  

And ∠ ABD  =  ∠  ACE  )

From equation 1 and 2 , we get

∠ CBD  =  ∠  BCE  ---------- ( 3 )

Now in ∆ CBD and ∆ BCE

BC =BC                               ( Common side )

∠ CBD  =  ∠  BCE           ( From equation 3 )

And BD=CE                                                  ( Given )

Hence

∆ CBD ≅∆ BCE                     ( By SAS  rule )

So,BE =DC                                                                              (  By CPCT  )  ( Hence proved )

Step-by-step explanation:

Answered by anjudpatil14
4

Answer:

plzzz see the pictures and mark answer as brainliest answer

Step-by-step explanation:

Given:△ABD and △ACE are equilateral triangle.

∴ In △ABD

∠BAD=∠ABD=∠BDA=60

.....(1)

In △ACE

∠AEC=∠ACE=∠ACE=60

.....(2)

To prove:∠CAD=∠BAE

Proof:∠CAD=∠BAC+∠DAB

⇒∠CAD=∠BAC+60

using (1)

⇒∠CAD=∠BAC+∠CAE using (2)

⇒∠CAD=∠BAE

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