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  • CDE is an equilateral triangle formed on a side CD of a square ABCD (Figure). Show that Triangle ADE congurent Triangle BCE.

CDE is an equilateral triangle formed on a side CD of a square ABCD (Figure). Show that \triangle ADE \cong \triangle BCE.

Answers (2)

Given, that CDE is an equilateral triangle

Then DE = EC = DC

and \angleCDE = \angleDEC = \angleECD = 60°

To prove: \triangleADE \cong \triangleBCE

Proof :

ABCD is a square then,

AB = BC = CD = AD

\angleA = \angleD = \angleB = \angleC = 90°

\angleADE = \angleADC + \angleCDE = 90° + 60°          

and \angleBCE = \angleBCD + \angleDCE = 90° + 60°

Then \angleADE = \angleBCE

In \triangleADE and \triangleBCE

AD = BC                     (Given)

DE = EC                      (Given)

\angleADE = \angleBCE          (from above)

By SAS criterion of congruence

\triangleADE \cong \triangleBCE

Hence proved.

Posted by

infoexpert24

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Given, CDE is an equilateral triangle

Then DE = EC = DC

and \angleCDE = \angleDEC = \angleECD = 60°

To prove : \triangleADE \cong \triangleBCE

Proof :

ABCD is a square then,

AB = BC = CD = AD

\angleA = \angleD = \angleB = \angleC = 90°

\angleADE = \angleADC + \angleCDE = 90° + 60°          

and \angleBCE = \angleBCD + \angleDCE = 90° + 60°

Then \angleADE = \angleBCE

In \triangleADE and \triangleBCE

AD = BC                     (Given)

DE = EC                      (Given)

\angleADE = \angleBCE          (from above)

By SAS criterion of congruence

\triangleADE \cong \triangleBCE

Hence proved.

Posted by

infoexpert24

View full answer

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