Question

In an equilateral triangle ABC, AD is drawn perpendicular to BC meeting BC in A. Prove that AD^2=3BD^2.

Answer 1

I have attached the solution here...you can check and if you like then please give an upvote cause this is the first answer i am uploading :)

Answer 2

Answer:

Hence proved that $\mathrm { AD } ^ { 2 } = 3 \mathrm { BD } ^ { 2 }$ in ∆ABC

To prove:

The value of $\mathrm { AD } ^ { 2 } = 3 \mathrm { BD } ^ { 2 }$

Solution:

Given that ∆ ABC and AD is perpendicular to BC i.e., AD ⊥ BC

Proof: Consider triangle ABD

$\mathrm { AB } ^ { 2 } = \mathrm { AD } ^ { 2 } + \mathrm { BD } ^ { 2 }$

(by using Pythagoras theorem)

$\mathrm { BC } ^ { 2 } = \mathrm { AD } ^ { 2 } + \mathrm { BD } ^ { 2 } ( \mathrm { AB } = \mathrm { BC } )$

Since all the sides in an equilateral triangle are equal.

Here,  

$B C ^ { 2 } = B D ^ { 2 } + D C ^ { 2 }$

Here, the perpendicular bisector bisects the line BC equally. Hence,

$BD=DC$

Thus,

$\begin{array} { c } { B C ^ { 2 } = B D ^ { 2 } + B D ^ { 2 } = 2 B D ^ { 2 } } \\\\ { ( 2 B D ) ^ { 2 } = A D ^ { 2 } + B D ^ { 2 } } \\\\ { 4 B D ^ { 2 } = A D ^ { 2 } + B D ^ { 2 } } \\\\ { 4 B D ^ { 2 } - B D ^ { 2 } = A D ^ { 2 } } \\\\ { 3 B D ^ { 2 } = A D ^ { 2 } } \end{array}$

Hence proved that $3 \mathrm { BD } ^ { 2 } = \mathrm { AD } ^ { 2 }$ in the given triangle ABC.