Question

State and prove the Pythagoras theorem.

Answer 1

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 $\mathrm{\angle }\text{ABC is 9}{\text{0}}^0$.
Construction: Draw a perpendicular BD on AC i.e. BD $\mathrm{\perp }$ AC.

In $\mathrm{\Delta }\text{ABD and }\mathrm{\Delta }\text{ABC }$ we have,
$\mathrm{\angle }\text{BAD = }\mathrm{\angle }\text{BAC }$i.e. $\mathrm{\angle }\text{A}$ is common in both triangles.
$\mathrm{\angle }\text{ABC = }\mathrm{\angle }\text{ADB = 9}{\text{0}}^0$
Therefore $\mathrm{\Delta }\text{ABC}\sim \mathrm{\Delta }\text{ABD }$( By AA similarity i.e. angle-angle similarity)
So,$$\begin{gathered} \Rightarrow {\displaystyle \frac{\text{AD}}{\text{AB}}}={\displaystyle \frac{\text{AB}}{\text{AC}}} \\ \Rightarrow \text{A}{\text{B}}^2\text{ = AD}\times \text{AC }...\text{(1)} \end{gathered}$$
In $\mathrm{\Delta }\text{BDC and }\mathrm{\Delta }\text{ABC }$ we have,
$\mathrm{\angle }\text{BCD = }\mathrm{\angle }\text{BCA }$i.e. $\mathrm{\angle }\text{C}$ is common in both triangles.
$\mathrm{\angle }\text{ABC = }\mathrm{\angle }\text{ADC = 9}{\text{0}}^0$
Therefore $\mathrm{\Delta }\text{ABC}\sim \mathrm{\Delta }\text{BDC }$( By AA similarity i.e. angle-angle similarity)
So,$$\begin{gathered} \Rightarrow {\displaystyle \frac{\text{DC}}{\text{BC}}}={\displaystyle \frac{\text{BC}}{\text{AC}}} \\ \Rightarrow \text{B}{\text{C}}^2\text{ = AC}\times \text{DC }...\text{(2)} \end{gathered}$$
Adding equation (1) and (2) , we get
$$\begin{gathered} \Rightarrow \text{A}{\text{B}}^2\text{ + B}{\text{C}}^2\text{ = AD}\times \text{AC + AC}\times \text{ DC} \\ \Rightarrow \text{A}{\text{B}}^2\text{ + B}{\text{C}}^2\text{ = AC(AD + DC)} \\ \Rightarrow \text{A}{\text{B}}^2\text{ + B}{\text{C}}^2\text{ = AC(AC)} \\ \Rightarrow \text{A}{\text{B}}^2\text{ + B}{\text{C}}^2\text{ = A}{\text{C}}^2 \end{gathered}$$
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.