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

Answer

Verified

365.7k+ 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.

Last updated date: 02nd Oct 2023

â€¢

Total views: 365.7k

â€¢

Views today: 4.65k

Recently Updated Pages

What do you mean by public facilities

Paragraph on Friendship

Slogan on Noise Pollution

Disadvantages of Advertising

Prepare a Pocket Guide on First Aid for your School

10 Slogans on Save the Tiger

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

One cusec is equal to how many liters class 8 maths CBSE

The equation xxx + 2 is satisfied when x is equal to class 10 maths CBSE

What is the color of ferrous sulphate crystals? How does this color change after heating? Name the products formed on strongly heating ferrous sulphate crystals. What type of chemical reaction occurs in this type of change.

Give 10 examples for herbs , shrubs , climbers , creepers

Change the following sentences into negative and interrogative class 10 english CBSE