Question

the side AB and AC of a triangle ABC is equilateral triangle ABD and ACE are drawn .
PROVE:-
(I) <CAD=<BAE
(II)CD=BE ​

Answer 1

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$\implies$Given :

Triangle ABC in which ABD and ACE are equilateral triangles

$\implies$To prove :

i ) < CAD = < BAE

ii ) CD = BE

$\implies$Construction :

Join CD and BE

$\implies$Proof :

let < BAC be x

ABD is equilateral triangle so each angle is 60°

<DAB + < BAC = < CAD

$\implies$60° + x° = <CAD ................. ( i )

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

< EAC + < BAC = < BAE

$\implies$60 + x° = <BAE ..................... ( i )

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

$\implies$2 ) To prove is CD = BE

Comparing Triangle DAC and triangle EAB

$\implies$< DAC = < BAE ( From i and ii )

$\implies$AD = AB ( given )

$\implies$AC = AE ( given )

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

$\implies$so , CB = BE by C.P.C.T.C.E

$\implies$$\implies$Hence proved !

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