Math, asked by adityamakar23, 5 months ago

In an equilateral triangle ABC, the side BC is trisected at D. Prove that 9AD2 = 7AB2. ​

Answers

Answered by bhaktihbalwadkar
0

Answer:

Let side =a

Draw AE⊥BC So, as triangle ABC is equilateral.

BE=

2

a

and as BD=

3

a

DE=BE−BD=

6

a

In △ADE by Pythagoras theorem

AD

2

=AE

2

+DE

2

=(

2

3

a)

2

+

36

a

2

=

9

7a

2

=

9

7AB

2

⇒9AD

2

=7AB

2

Answered by singhdipanshu2707200
2

Answer:

Check your answer please

Attachments:
Similar questions