Question

On the sides AB and AC of a triangle ABC are equilateral triangles ABD , ACE are drawn . Prove that <CAD = <BAE , ii ) CD = BE ​

Answer 1

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Given :

Triangle ABC in which ABD and ACE are equilateral triangles

To prove :

< CAD = < BAE

ii ) CD = BE

Construction :

Join CD and BE

Proof :

let < BAC be x

ABD is equilateral triangle so each angle is 60°

<DAB + < BAC = < CAD

60° + x° = <CAD ................. ( i )

Triangle ACE is equilateral triangle , so each angle is 60 °

< EAC + < BAC = < BAE

60 + x° = <BAE ..................... ( i )

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From i and ii we get <CAD = < BAE ( proved 1 )

2 ) To prove is CD = BE

Comparing Triangle DAC and triangle EAB

< DAC = < BAE ( From i and ii )

AD = AB ( given )

AC = AE ( given )

so now , Triangle DAC is congurent to triangle EAB by the rule SIDE . ANGLE . SIDE ( S.A.S )

so , CB = BE by C.P.C.T.C.E

Hence proved !