8. Upon the sides AB, AC of a ABC are
described equilateral triangles ABD and
ACE with their vertices D and E outside
the ABC. Prove that BE DC.
and
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 )
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Ankitha Vyas
Grade 9
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