in the adjacent figure triangle CDE is an equilateral triangle formed on a side cd of a square ABCD show that triangle ABC congruent triangle BCE
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Answered by
57
Hello Mate!
Error correction : It will be "show that ∆ADE ~ ∆BCE"
Given : ∆CDE is equilateral ∆ on side CD of square ABCD.
To prove : ∆ADE ~ ∆BCE.
Proof : In ∆ADE and ∆BCE
Since sides of square are equal
AD = BC _(i)
Since side of equilateral ∆ are equal
DE = CE _(ii)
Now, < ADC = < BCD = 90°
< CDE = < DCE = 60°
Adding borh equation we get,
< ADE = BCE = 150° _(iii)
Hence by (i), (ii) and (iii) we prove that,
∆ADE ~ BCE ( by SAS congruency )
Therefore, ar(∆ADE) = ar(∆BCE)
Q.E.D
Have great future ahead!
Error correction : It will be "show that ∆ADE ~ ∆BCE"
Given : ∆CDE is equilateral ∆ on side CD of square ABCD.
To prove : ∆ADE ~ ∆BCE.
Proof : In ∆ADE and ∆BCE
Since sides of square are equal
AD = BC _(i)
Since side of equilateral ∆ are equal
DE = CE _(ii)
Now, < ADC = < BCD = 90°
< CDE = < DCE = 60°
Adding borh equation we get,
< ADE = BCE = 150° _(iii)
Hence by (i), (ii) and (iii) we prove that,
∆ADE ~ BCE ( by SAS congruency )
Therefore, ar(∆ADE) = ar(∆BCE)
Q.E.D
Have great future ahead!
Anonymous:
Nice answer :)
Answered by
28
We have to prove triangle ADE congruent to BCE
As ABCD is a square
So AB= BC=DC= DA= x
As DCE is an equilateral triangle
So DC= CE= DE
As DC is also side of square = x
So DC= CE= DE= AB= BC= DA= x
So in triangle ADE
AD= DE
hence it becomes isosceles
Also
<D= 90 +60 = 150
So DAE= DEA = 15 ( As 180 - 2 DAE= 150)
Similarly in BCE
BC= CE
So it also becomes isosceles
Also <C= 90+60= 150
<CBE= <CEB= 15
So both ADE and BCE becomes similar isosceles triangles
As AD= BC= DE= CE
and <C= <D = 150
<DAE= <DEA= <CBE= <CEB = 15
hence ADE and BCE are congruent
✌✌✌✌Dr.Dhruv✌✌✌✌✌
As ABCD is a square
So AB= BC=DC= DA= x
As DCE is an equilateral triangle
So DC= CE= DE
As DC is also side of square = x
So DC= CE= DE= AB= BC= DA= x
So in triangle ADE
AD= DE
hence it becomes isosceles
Also
<D= 90 +60 = 150
So DAE= DEA = 15 ( As 180 - 2 DAE= 150)
Similarly in BCE
BC= CE
So it also becomes isosceles
Also <C= 90+60= 150
<CBE= <CEB= 15
So both ADE and BCE becomes similar isosceles triangles
As AD= BC= DE= CE
and <C= <D = 150
<DAE= <DEA= <CBE= <CEB = 15
hence ADE and BCE are congruent
✌✌✌✌Dr.Dhruv✌✌✌✌✌
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