Math, asked by vanshita14, 1 year ago

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 ShuchiRecites
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!

Anonymous: Nice answer :)
ShuchiRecites: Thanks asp39 sis
Answered by Anonymous
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✌✌✌✌✌
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Anonymous: Nice answer :)
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