# State and prove the Pythagoras theorem.

Last updated date: 22nd Mar 2023

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Hint: Draw a perpendicular on AC from B and use angle-angle similarity to prove the theorem.

Complete step-by-step answer:

Pythagoras theorem states that “ In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

The sides of the right-angled triangle are called base, perpendicular and hypotenuse .

According to Pythagoras theorem ,

${({\text{AC)}}^2} = {({\text{AB)}}^2}{\text{ + (BC}}{{\text{)}}^2}$

Proof:

Given, a triangle ABC in which $\angle {\text{ABC is 9}}{{\text{0}}^0}$.

Construction: Draw a perpendicular BD on AC i.e. BD $ \bot $ AC.

In $\Delta {\text{ABD and }}\Delta {\text{ABC }}$ we have,

$\angle {\text{BAD = }}\angle {\text{BAC }}$i.e. $\angle {\text{A}}$ is common in both triangles.

$\angle {\text{ABC = }}\angle {\text{ADB = 9}}{{\text{0}}^0}$

Therefore $\Delta {\text{ABC}} \sim \Delta {\text{ABD }}$( By AA similarity i.e. angle-angle similarity)

So,$

\Rightarrow \dfrac{{{\text{AD}}}}{{{\text{AB}}}} = \dfrac{{{\text{AB}}}}{{{\text{AC}}}} \\

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ = AD}} \times {\text{AC }}...{\text{(1)}} \\

$

In $\Delta {\text{BDC and }}\Delta {\text{ABC }}$ we have,

$\angle {\text{BCD = }}\angle {\text{BCA }}$i.e. $\angle {\text{C}}$ is common in both triangles.

$\angle {\text{ABC = }}\angle {\text{ADC = 9}}{{\text{0}}^0}$

Therefore $\Delta {\text{ABC}} \sim \Delta {\text{BDC }}$( By AA similarity i.e. angle-angle similarity)

So,$

\Rightarrow \dfrac{{{\text{DC}}}}{{{\text{BC}}}} = \dfrac{{{\text{BC}}}}{{{\text{AC}}}} \\

\Rightarrow {\text{B}}{{\text{C}}^2}{\text{ = AC}} \times {\text{DC }}...{\text{(2)}} \\

$

Adding equation (1) and (2) , we get

$

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}{\text{ = AD}} \times {\text{AC + AC}} \times {\text{ DC}} \\

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}{\text{ = AC(AD + DC)}} \\

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}{\text{ = AC(AC)}} \\

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}{\text{ = A}}{{\text{C}}^2} \\

$

Hence, proved.

Note: In a right angled triangle , hypotenuse is the longest side of the triangle and is opposite to the right angle i.e. ${90^0}$. By drawing a perpendicular from point B and dividing the triangle ABC into 2 parts and using angle-angle similarity to prove the Pythagoras theorem.

Complete step-by-step answer:

Pythagoras theorem states that “ In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

The sides of the right-angled triangle are called base, perpendicular and hypotenuse .

According to Pythagoras theorem ,

${({\text{AC)}}^2} = {({\text{AB)}}^2}{\text{ + (BC}}{{\text{)}}^2}$

Proof:

Given, a triangle ABC in which $\angle {\text{ABC is 9}}{{\text{0}}^0}$.

Construction: Draw a perpendicular BD on AC i.e. BD $ \bot $ AC.

In $\Delta {\text{ABD and }}\Delta {\text{ABC }}$ we have,

$\angle {\text{BAD = }}\angle {\text{BAC }}$i.e. $\angle {\text{A}}$ is common in both triangles.

$\angle {\text{ABC = }}\angle {\text{ADB = 9}}{{\text{0}}^0}$

Therefore $\Delta {\text{ABC}} \sim \Delta {\text{ABD }}$( By AA similarity i.e. angle-angle similarity)

So,$

\Rightarrow \dfrac{{{\text{AD}}}}{{{\text{AB}}}} = \dfrac{{{\text{AB}}}}{{{\text{AC}}}} \\

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ = AD}} \times {\text{AC }}...{\text{(1)}} \\

$

In $\Delta {\text{BDC and }}\Delta {\text{ABC }}$ we have,

$\angle {\text{BCD = }}\angle {\text{BCA }}$i.e. $\angle {\text{C}}$ is common in both triangles.

$\angle {\text{ABC = }}\angle {\text{ADC = 9}}{{\text{0}}^0}$

Therefore $\Delta {\text{ABC}} \sim \Delta {\text{BDC }}$( By AA similarity i.e. angle-angle similarity)

So,$

\Rightarrow \dfrac{{{\text{DC}}}}{{{\text{BC}}}} = \dfrac{{{\text{BC}}}}{{{\text{AC}}}} \\

\Rightarrow {\text{B}}{{\text{C}}^2}{\text{ = AC}} \times {\text{DC }}...{\text{(2)}} \\

$

Adding equation (1) and (2) , we get

$

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}{\text{ = AD}} \times {\text{AC + AC}} \times {\text{ DC}} \\

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}{\text{ = AC(AD + DC)}} \\

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}{\text{ = AC(AC)}} \\

\Rightarrow {\text{A}}{{\text{B}}^2}{\text{ + B}}{{\text{C}}^2}{\text{ = A}}{{\text{C}}^2} \\

$

Hence, proved.

Note: In a right angled triangle , hypotenuse is the longest side of the triangle and is opposite to the right angle i.e. ${90^0}$. By drawing a perpendicular from point B and dividing the triangle ABC into 2 parts and using angle-angle similarity to prove the Pythagoras theorem.

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