Question

In the given figure, equilateral ΔABD and ΔACE are drawn on the sides of a ΔABC.
Prove that CD = BE.

Answer 1

DIAGRAM :-

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GIVEN :-

• Equilateral triangle ABD and ACE are drawn on the sides of triangle ABC.

TO PROVE :-

• CD = BE

SOLUTION :-

▪︎To prove CD = BE we have to prove first $\angle$ CAD = $\angle$ BAE

Now, $\angle$ CAD = $\angle$ CAB + $\angle$ BAD

=> $\angle$ CAD = $\angle$ CAB + 60° [ $\angle$ BAD = 60° ]

=> $\angle$ CAD = $\angle$ CAB + CAE [ CAE = 60° ]

=> $\angle$ CAD = $\angle$ BAE [ $\angle$ CAB + $\angle$ CAE = $\angle$ BAE

Hence proved

Now in triangle CAD and triangle BAE ,

=> BA = AD

In $\angle$ CAD = $\angle$ BAE

=> AC = AE

Hence triangle CAD ≅ triangle BAE .

so, CD = BE [CPCT]

HENCE PROVED

Answer 2

Answer:

GIVEN :-

Equilateral triangle ABD and ACE are drawn on the sides of triangle ABC.

TO PROVE :-

CD = BE

SOLUTION :-

▪︎To prove CD = BE we have to prove first \angle∠ CAD = \angle∠ BAE

Now, \angle∠ CAD = \angle∠ CAB + \angle∠ BAD

=> \angle∠ CAD = \angle∠ CAB + 60° [ \angle∠ BAD = 60° ]

=> \angle∠ CAD = \angle∠ CAB + CAE [ CAE = 60° ]

=> \angle∠ CAD = \angle∠ BAE [ \angle∠ CAB + \angle∠ CAE = \angle∠ BAE

Hence proved

Now in triangle CAD and triangle BAE ,

=> BA = AD

In \angle∠ CAD = \angle∠ BAE

=> AC = AE

Hence triangle CAD ≅ triangle BAE .

so, CD = BE [CPCT]

HENCE PROVED