Question

In an equilateral triangle ABC, D is a point on side BC such that 1/3 BC. prove that 9 AD^ =7 AB ^

Answer 1

hope this helps you....

Answer 2

Answer

Let the sides of eq. ∆ be a cm.

In ∆ ADE,

AD2= AE2 + DE2........ (1)

In ∆ AEC,

AC2= AE2 + EC2

AE2= AC2 - EC2

= a2 - (a/ 2)^2. (AE is perp to BC)

= a^2-a^2/4

= 3a^2/4................. (2)

DE = BE - BD

= a/2 - a/3

= a/6......................... (3)

Using 2 and 3 in 1,

AD2= AC2 + DE2

AD2 = 3a^2/4 + (a/6)^2

AD2 = 3a^2/4 + a^2/36

AD2 =(27a^2 + a^2)/2

AD2 = 28 a2/36

AD2= 7a^2/9

9AD2 = 7a2

9AD2 = 7AB2........

Hence proved...

Hope u'll understand......