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.
Answers
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:
Answer:
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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