In an equilateral triangle ABC, D is a point on side BC such that ${\text{BD}} = \dfrac{1}{3}\left( {{\text{BC}}} \right)$. Prove that $9{\left( {{\text{AD}}} \right)^2} = 7{\left( {{\text{AB}}} \right)^2}$.
.

Last updated date: 29th Mar 2023
•
Total views: 310.8k
•
Views today: 2.88k
Answer
310.8k+ views
Hint- Here, we will be using Pythagoras theorem and will be converting all other sides in terms of sides AD and AB which are present in the equation which needs to be proved.
To prove- $9{\left( {{\text{AD}}} \right)^2} = 7{\left( {{\text{AB}}} \right)^2}$
Let us draw an equilateral $\vartriangle {\text{ABC}}$ with sides AB, BC and AC equal to each other. Also, draw a line from the vertex A which will meet the side BC at point D such that ${\text{BD}} = \dfrac{1}{3}\left( {{\text{BC}}} \right)$. Also, draw a perpendicular from the vertex A on the side BC which divides line BC into two equal parts i.e., ${\text{BE}} = {\text{CE}} = \dfrac{{{\text{BC}}}}{2}$
As, ${\text{BD}} = \dfrac{1}{3}\left( {{\text{BC}}} \right){\text{ }} \to {\text{(1)}}$
Here, ${\text{AB}} = {\text{BC}} = {\text{AC}}$ and ${\text{DE}} = \left( {{\text{BE}} - {\text{BD}}} \right)$
As we know that, Pythagoras theorem states that the square of the hypotenuse is equal to the sum of the squares of perpendicular and base in a right angled triangle.
In right angled $\vartriangle {\text{ADE}}$,
Using Pythagoras theorem \[{\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AE}}} \right)^2} + {\left( {{\text{DE}}} \right)^2} \Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AE}}} \right)^2} + {\left( {{\text{BE}} - {\text{BD}}} \right)^2}{\text{ }} \to {\text{(2)}}\]
In right angled $\vartriangle {\text{ABE}}$,
Using Pythagoras theorem \[{\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AE}}} \right)^2} + {\left( {{\text{BE}}} \right)^2} \Rightarrow {\left( {{\text{AE}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - {\left( {{\text{BE}}} \right)^2} \Rightarrow {\left( {{\text{AE}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - {\left( {\dfrac{{{\text{BC}}}}{2}} \right)^2}{\text{ }} \to {\text{(3)}}\]
Now using equation (3) in equation (2), we get
\[ \Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - {\left( {\dfrac{{{\text{BC}}}}{2}} \right)^2} + {\left( {{\text{BE}} - {\text{BD}}} \right)^2}{\text{ }}\]
Using ${\text{BE}} = \dfrac{{{\text{BC}}}}{2}$ and ${\text{BD}} = \dfrac{1}{3}\left( {{\text{BC}}} \right)$ , the above equation becomes
\[
\Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - {\left( {\dfrac{{{\text{BC}}}}{2}} \right)^2} + {\left( {\dfrac{{{\text{BC}}}}{2} - \dfrac{{{\text{BC}}}}{3}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{4}} \right] + {\left( {\dfrac{{{\text{3BC}} - {\text{2BC}}}}{6}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{4}} \right] + \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{{36}}} \right] \\
\Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} + \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{{36}} - \dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{4}} \right] = {\left( {{\text{AB}}} \right)^2} + \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2} - 9{{\left( {{\text{BC}}} \right)}^2}}}{{36}}} \right] = {\left( {{\text{AB}}} \right)^2} + \left[ {\dfrac{{ - 8{{\left( {{\text{BC}}} \right)}^2}}}{{36}}} \right] = {\left( {{\text{AB}}} \right)^2} - \left[ {\dfrac{{2{{\left( {{\text{BC}}} \right)}^2}}}{9}} \right] \\
\]
Also, ${\text{AB}} = {\text{BC}} = {\text{AC}} \Rightarrow {\text{BC}} = {\text{AB}}$so the above equation becomes
\[
\Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - \left[ {\dfrac{{2{{\left( {{\text{BC}}} \right)}^2}}}{9}} \right] = {\left( {{\text{AB}}} \right)^2} - \dfrac{{2{{\left( {{\text{AB}}} \right)}^2}}}{9} = \dfrac{{9{{\left( {{\text{AB}}} \right)}^2} - 2{{\left( {{\text{AB}}} \right)}^2}}}{9} = \dfrac{{7{{\left( {{\text{AB}}} \right)}^2}}}{9} \\
\Rightarrow 9{\left( {{\text{AD}}} \right)^2} = 7{\left( {{\text{AB}}} \right)^2} \\
\]
The above equation is the equation we needed to prove. Hence it is proved.
Note- In any right angled triangle, the side opposite to the right angle is known as hypotenuse. In this particular problem in order to obtain the LHS of the equation which needs to be proved, Pythagoras theorem is applied in the right triangle ADE and then all the other sides except AD are converted in terms of side A
To prove- $9{\left( {{\text{AD}}} \right)^2} = 7{\left( {{\text{AB}}} \right)^2}$
Let us draw an equilateral $\vartriangle {\text{ABC}}$ with sides AB, BC and AC equal to each other. Also, draw a line from the vertex A which will meet the side BC at point D such that ${\text{BD}} = \dfrac{1}{3}\left( {{\text{BC}}} \right)$. Also, draw a perpendicular from the vertex A on the side BC which divides line BC into two equal parts i.e., ${\text{BE}} = {\text{CE}} = \dfrac{{{\text{BC}}}}{2}$
As, ${\text{BD}} = \dfrac{1}{3}\left( {{\text{BC}}} \right){\text{ }} \to {\text{(1)}}$
Here, ${\text{AB}} = {\text{BC}} = {\text{AC}}$ and ${\text{DE}} = \left( {{\text{BE}} - {\text{BD}}} \right)$
As we know that, Pythagoras theorem states that the square of the hypotenuse is equal to the sum of the squares of perpendicular and base in a right angled triangle.
In right angled $\vartriangle {\text{ADE}}$,
Using Pythagoras theorem \[{\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AE}}} \right)^2} + {\left( {{\text{DE}}} \right)^2} \Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AE}}} \right)^2} + {\left( {{\text{BE}} - {\text{BD}}} \right)^2}{\text{ }} \to {\text{(2)}}\]
In right angled $\vartriangle {\text{ABE}}$,
Using Pythagoras theorem \[{\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AE}}} \right)^2} + {\left( {{\text{BE}}} \right)^2} \Rightarrow {\left( {{\text{AE}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - {\left( {{\text{BE}}} \right)^2} \Rightarrow {\left( {{\text{AE}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - {\left( {\dfrac{{{\text{BC}}}}{2}} \right)^2}{\text{ }} \to {\text{(3)}}\]
Now using equation (3) in equation (2), we get
\[ \Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - {\left( {\dfrac{{{\text{BC}}}}{2}} \right)^2} + {\left( {{\text{BE}} - {\text{BD}}} \right)^2}{\text{ }}\]
Using ${\text{BE}} = \dfrac{{{\text{BC}}}}{2}$ and ${\text{BD}} = \dfrac{1}{3}\left( {{\text{BC}}} \right)$ , the above equation becomes
\[
\Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - {\left( {\dfrac{{{\text{BC}}}}{2}} \right)^2} + {\left( {\dfrac{{{\text{BC}}}}{2} - \dfrac{{{\text{BC}}}}{3}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{4}} \right] + {\left( {\dfrac{{{\text{3BC}} - {\text{2BC}}}}{6}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{4}} \right] + \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{{36}}} \right] \\
\Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} + \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{{36}} - \dfrac{{{{\left( {{\text{BC}}} \right)}^2}}}{4}} \right] = {\left( {{\text{AB}}} \right)^2} + \left[ {\dfrac{{{{\left( {{\text{BC}}} \right)}^2} - 9{{\left( {{\text{BC}}} \right)}^2}}}{{36}}} \right] = {\left( {{\text{AB}}} \right)^2} + \left[ {\dfrac{{ - 8{{\left( {{\text{BC}}} \right)}^2}}}{{36}}} \right] = {\left( {{\text{AB}}} \right)^2} - \left[ {\dfrac{{2{{\left( {{\text{BC}}} \right)}^2}}}{9}} \right] \\
\]
Also, ${\text{AB}} = {\text{BC}} = {\text{AC}} \Rightarrow {\text{BC}} = {\text{AB}}$so the above equation becomes
\[
\Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AB}}} \right)^2} - \left[ {\dfrac{{2{{\left( {{\text{BC}}} \right)}^2}}}{9}} \right] = {\left( {{\text{AB}}} \right)^2} - \dfrac{{2{{\left( {{\text{AB}}} \right)}^2}}}{9} = \dfrac{{9{{\left( {{\text{AB}}} \right)}^2} - 2{{\left( {{\text{AB}}} \right)}^2}}}{9} = \dfrac{{7{{\left( {{\text{AB}}} \right)}^2}}}{9} \\
\Rightarrow 9{\left( {{\text{AD}}} \right)^2} = 7{\left( {{\text{AB}}} \right)^2} \\
\]
The above equation is the equation we needed to prove. Hence it is proved.
Note- In any right angled triangle, the side opposite to the right angle is known as hypotenuse. In this particular problem in order to obtain the LHS of the equation which needs to be proved, Pythagoras theorem is applied in the right triangle ADE and then all the other sides except AD are converted in terms of side A
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
