Question

In an equilateral triangle ABC D is a
point on side BC such that
BD = 1/3 BC Prove that
9AD² = 7AB²​

Answer 1

Here’s your answer:-

Answer 2

Answer:

Step-by-step explanation:

given: BD=1/3BC

p.t : 9 AD² = 7 AB²

PROOF : IN TRIANGLE ADE

AD²=AE² +DE²

AB²=AE² + EB²

AD² - AB² = DE² - EB²

AD² - AB² = [ BE² -BD²] -[BE²]

                 =[ 1/2 BC²-1/3 BC² ] -[ 1/2 BC² ]                        [ BE= 1/2 BC ]

                 = [ 1/6 BC ]²  - [ 1/2 BC ] ²

                  = 1/36 BC ²- 1/4 BC²

                   = 1 BC² -9 BC² /36

                 =  -8 BC²/36

AD² - AB² = -8 BC² /36

AD²-BC = -8 BC²/36

AD² = -8 BC² / 36 + BC /1    =    28 AB/36

AD²= 7 AB² / 9

9 AD² = 7 AB²

                                     HENCE PROVED