Question

State and prove the Pythagoras theorem.

Answer 1

Hint: In this question we have to give the statement of the Pythagoras theorem and prove it step by step using a trigonometric method. The Pythagoras theorem is only applicable for a right-angled triangle. A right-angled triangle is a triangle which has one angle at 90 degrees. The Pythagoras theorem gives the relationship between the height, base and hypotenuse of the right-angled triangle.

Complete step-by-step answer:
Given:
Let us assume the triangle $ \Delta ABC $ is a right-angle triangle and the angle $ \angle B = 90^\circ $ . Side AB is the height, BC is the base and AC is the hypotenuse of the triangle.
Now drawing a perpendicular BD on the hypotenuse AC of the triangle, we have now two triangles $ \Delta ADB $ and $ \Delta BDC $ .

Since both the triangles $ \Delta ADB $ and $ \Delta BDC $ have one angle at 90 degrees and one angle common with $ \Delta ABC $ so we can say that both the triangles $ \Delta ADB $ and $ \Delta BDC $ are similar triangles with $ \Delta ABC $ .
So, for a similar triangle the property is given as – “for two similar triangles the corresponding sides of both the similar triangles have the constant ratio”.
Now using the similar triangles property, we have –
 $ \Delta ADB \simeq \Delta ABC $
\[\Rightarrow \dfrac{{AD}}{{AB}} = \dfrac{{AB}}{{AC}}\]
Solving this we get,
\[\Rightarrow A{B^{2\;}} = {\rm{ }}AD\; \times {\rm{ }}AC\]
This is our first equation.
Similarly, for the similar triangles \[\Delta BDC\] and \[\Delta ABC\] using the similar triangle property, we have,
\[\Rightarrow \Delta BDC \simeq \Delta ABC\]
\[\Rightarrow \dfrac{{CD}}{{BC}} = \dfrac{{BC}}{{AC}}\]
Solving this we get,
\[\;B{C^2} = {\rm{ }}CD{\rm{ }} \times {\rm{ }}AC\]
This is our second equation.
Now adding the first and second equations, we get,
\[
\Rightarrow {A{B^{2\;}} + {\rm{ }}B{C^{2\;}} = {\rm{ }}AD\; \times AC{\rm{ }} + \;CD \times AC}\\
\Rightarrow {A{B^{2\;}} + {\rm{ }}B{C^{2\;}} = {\rm{ }}AC{\rm{ }}\left( {AD{\rm{ }} + {\rm{ }}CD} \right)}
\]
Since we know that from the figure
\[AD{\rm{ }} + {\rm{ }}CD{\rm{ }} = {\rm{ }}AC\]
Substituting this value into the equation we get,
\[\Rightarrow A{C^2}\; = {\rm{ }}A{B^2}\; + {\rm{ }}B{C^2}\]
It means that the square of the hypotenuse is always equal to the sum of the square of the base and height of the right-angled triangle. This is the statement for the Pythagoras theorem.
Hence, the Pythagorean theorem is proved.

Note: It should be noted that the Pythagoras theorem can be applied to only right-angled triangles, it means that if any triangle which has one inner angle of 90 degrees then we can apply the Pythagoras theorem.