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Maths-

General

Easy

Question

In an equilateral triangle ABC,D is a point on the side BC , such that $BD equals 1 third BC$ . Prove that
$9 AD squared equals 7 AB squared$

The correct answer is: Hence proved

Solution :-
Aim :- Prove that $9 A D squared equals 7 A B squared$
Hint :- let the side of equilateral triangle be x , then we get height of equilateral
triangle as $left parenthesis x square root of 3 right parenthesis divided by 2$ .Now , Find length of DE and apply pythagoras theorem in
Right angle triangle ADE to find length of AD in terms of AB.
Explanation(proof ) :-
In equilateral triangle all sides are equal
Let , AB = BC = CA = x
Given BD = ⅓ BC = x/3
In equilateral triangle ,the altitudes and medians coincide
so,AE is median also i.e E is midpoint of BC
BE = x/2
From diagram ,DE = BE - BD $equals x over 2 minus x over 3 equals x over 6$
We get height of equilateral triangle when side x is $fraction numerator x square root of 3 over denominator 2 end fraction$
Applying pythagoras theorem on triangle ADE
We get $A D squared equals D E squared plus A E squared equals open parentheses x over 6 close parentheses squared plus open parentheses fraction numerator x square root of 3 over denominator 2 end fraction close parentheses squared$
$∣ A D squared equals x squared over 36 plus fraction numerator 3 x squared over denominator 4 end fraction equals fraction numerator 28 x squared over denominator 36 end fraction equals 7 over 0 x squared$

Substitute x by AB as x = AB
$A D squared equals 1 over 9 A B squared not stretchy rightwards double arrow 9 A D squared equals 7 A B squared$
Hence proved

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