Question

in fig. Dis point of line BC such that CD=2BC if triangle ABC is an equilateral triangle then prove that AD² = 7BC² ​

Answer 1

Given : D is point of line BC such that CD=2BC

triangle ABC is an equilateral triangle

To Find : prove that AD² = 7BC² ​

Solution:

Let say AB = BC = CD  = 2x

CD = 2BC = 2(2x)  = 4x

Draw  AE ⊥ BC

E  will be mid point of BC  as ABC is equilateral triangle

=> BE = CE = (1/2)BC = x

DE =  CD + CE =   x  + 4x  = 5x

AE² = AB²  - BE²

=> AE²  = (2x)²  - x²

=> AE²  = 3x²

AD²  =  AE²  + DE²

=> AD² = 3x²  + (5x)²

=> AD²  = 3x²  + 25x²

=> AD²  = 28x²

=> AD² = 7 * 4x²

=> AD² = 7(2x)²

=> AD² = 7 BC²

QED

Hence Proved

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