# In $\Delta {\text{ABC}}$, AD is perpendicular to BC. Prove that ${\left( {{\text{AB}}} \right)^2} + {\left( {{\text{CD}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} + {\left( {{\text{BD}}} \right)^2}$.

Last updated date: 19th Mar 2023

•

Total views: 308.1k

•

Views today: 7.86k

Answer

Verified

308.1k+ views

Hint- Here, we will be proceeding further with the help of Pythagoras theorem.

Clearly, from the figure we can see that there are total two right angled triangles $\Delta {\text{ABD}}$ and $\Delta {\text{ACD}}$.

As we know that the side opposite to the right angle in any right angled triangle is hypotenuse, the side opposite to the considered acute angle is perpendicular and the remaining side is base.

In right angled $\Delta {\text{ABD}}$, AB is a hypotenuse and in $\Delta {\text{ACD}}$, AC is a hypotenuse.

According to Pythagoras theorem, we know that

${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$

Therefore, in right angled $\Delta {\text{ABD}}$, ${\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{BD}}} \right)^2}{\text{ }} \to {\text{(1)}}$

and in right angled $\Delta {\text{ACD}}$, ${\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{CD}}} \right)^2} \Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} - {\left( {{\text{CD}}} \right)^2}{\text{ }} \to {\text{(2)}}$

Using equation (2) in equation (1), we get

${\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} - {\left( {{\text{CD}}} \right)^2} + {\left( {{\text{BD}}} \right)^2} \Rightarrow {\left( {{\text{AB}}} \right)^2} + {\left( {{\text{CD}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} + {\left( {{\text{BD}}} \right)^2}$

The above equation is what we needed to prove.

Note- In these types of problems, Pythagora's theorem is to be used keeping in mind the concept of a right triangle i.e., which side is to be taken hypotenuse. Also, in the above problem the choice between perpendicular and base is flexible since that choice depends upon the acute angle considered.

Clearly, from the figure we can see that there are total two right angled triangles $\Delta {\text{ABD}}$ and $\Delta {\text{ACD}}$.

As we know that the side opposite to the right angle in any right angled triangle is hypotenuse, the side opposite to the considered acute angle is perpendicular and the remaining side is base.

In right angled $\Delta {\text{ABD}}$, AB is a hypotenuse and in $\Delta {\text{ACD}}$, AC is a hypotenuse.

According to Pythagoras theorem, we know that

${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$

Therefore, in right angled $\Delta {\text{ABD}}$, ${\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{BD}}} \right)^2}{\text{ }} \to {\text{(1)}}$

and in right angled $\Delta {\text{ACD}}$, ${\left( {{\text{AC}}} \right)^2} = {\left( {{\text{AD}}} \right)^2} + {\left( {{\text{CD}}} \right)^2} \Rightarrow {\left( {{\text{AD}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} - {\left( {{\text{CD}}} \right)^2}{\text{ }} \to {\text{(2)}}$

Using equation (2) in equation (1), we get

${\left( {{\text{AB}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} - {\left( {{\text{CD}}} \right)^2} + {\left( {{\text{BD}}} \right)^2} \Rightarrow {\left( {{\text{AB}}} \right)^2} + {\left( {{\text{CD}}} \right)^2} = {\left( {{\text{AC}}} \right)^2} + {\left( {{\text{BD}}} \right)^2}$

The above equation is what we needed to prove.

Note- In these types of problems, Pythagora's theorem is to be used keeping in mind the concept of a right triangle i.e., which side is to be taken hypotenuse. Also, in the above problem the choice between perpendicular and base is flexible since that choice depends upon the acute angle considered.

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