Question

In fig. given below, ABCD is a square and triangleEDC is an equilateral triangle. Prove that :
•triangleADEQUATE congruent to triangle BCE
•AE=BE
•angleDAE=15

Answer 1

consider triangle PST and triangle QRT

PS = QR (side of a square)

TS = TR (sides of an equilateral triangle)

angle PST = angle QRT = 150 degrees ( 90 degrees + 60 degrees)

therefore SAS congruency triangle PST is congruent triangle QRT

PT = QT ((c.p.c.t.)

now, QR = RS

and RS = RT

so, QR = TR

this implies that, angle RTQ = angle RQT

angle RTQ + angle RQT + 150 degree = 180 degrees

2 angle RQT = 30 degrees

therefore angle RQT = 15 degrees


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Answer 2

Since ABCD is a square , so ADC = BCD = 90

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

now EDA = EDC + ADC = 150

ECB = ECD +BCD = 150

In triangles ADE and BCE,

AD = BC [sides of square are equal]

EDA = ECB [proved above]

ED = EC [Sides of equilateral triangle are equal]

Δ ADE is congruent to Δ BCE [ SAS ]

So, AE = BE [CPCT]

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

since CDE is an equilateral triangle, so CD = DE = EC ..............(2)

From (1) and (2), we have

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

In triangle DAE,

AD = DE [From (3)]

⇒ DEA = DAE [angles opposite to equal sides are equal]

In triangle DAE,

ADE + DEA + DAE = 180 [ANGLE SUM PROPERTY]

150 + 2DEA = 180

So, DEA = DAE = 15