Question

in the figure given, ABCD is a square and triangle DEC is an equilateral triangle. PROVE that 1.triangle ADE =~ triangle BCE, 2.AE=BE, 3.Angle=15°​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Hey friend here is your answer,

since ABCD is a square , so ADC = BCD = 90 °

since DEC is an equilateral triangle , EDC = ECD = 60°

Now, EDA=EDC+ ADC= 150°

in triangle ADE and BCE,

AD = BC ( sides of triangle)

EDA =ECB ( proved above)

ED= EC ( sides of equilateral triangle are equal)

therefore, triangle ADE is congruent to triangle BCE ( by C.P.C.T)

so, AE = BC ( by C.P.C.T)

since ABCD is a square, AB= BC= CD = AD......(1)

sil CDE is an equilateral triangle, so CD= DE=AC...(2)

from (1) and (2) we have,

AB=BC=AD=CD= DE=EC ........(3)

In triangle DAE,

AD= DE (from equation 3)

DEA = DAE ( angle opposite to equal sides are equal)

In triangle DAE,

ADE+DAE+DEA= 180° ( angle sum property)

150° +2DEA = 180°

so, DEA = DAE = 15°

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