Q. 154.3( 250 Votes )

# In an equilateral triangle ABC, D is a point on side BC such that Prove that 9 AD^{2 }= 7 AB^{2}

Answer :

The figure is given below:**Given: BD = BC/3**

**To Prove: 9 AD**

^{2}= 7 AB^{2}**Proof:**

Let the side of the equilateral triangle be

*a*, and AM be the altitude of ΔABC

BM = MC = BC/2 = a/2 [ Altitude of an equilateral triangle bisect the side]

And, then, in ΔABM, by pythagoras theorem we write,**Pythagoras Theorem : Square of the Hypotenuse equals to the sum of the squares of other two sides.**

AM^{2 }= AB^{2 }- BM^{2}

or AM^{2 }= a^{2 }- a^{2}/4

BD = a/3 [ BC = a]

DM = BM – BD

= a/2 – a/3

= a/6

^{ }

According to pythagoras theorem in a right angled triangle,

(hypotenuse)^{2 }= (altitude)^{2} + (base)^{2}

Applying Pythagoras theorem in ΔADM, we obtain

AD^{2} = AM^{2} + DM^{2 }

Now, a = AB or a^{2 } = AB^{2}

36 AD^{2 }= 28 AB^{2}

9 AD^{2} = 7 AB^{2 }Hence, Proved^{}

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