Question

State and prove the Pythagoras theorem.

Answer 1

Statement: In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: ABC is a triangle in which $\angle ABC={90}^{\circ }$

Construction: Draw $BD\perp AC$.

Proof:

In $△ADB$ and $△ABC$

$\angle A=\angle A$ [Common angle]

$\angle ADB=\angle ABC$ [Each ${90}^{\circ }$]

$△ADB\sim △ABC$ [$A-A$ Criteria]

So, $\frac{AD}{AB}=\frac{AB}{AC}$

Now, ${AB}^2=AD\times AC$ ..........(1)

Similarly,

${BC}^2=CD\times AC$ ..........(2)

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

${AB}^2+{BC}^2=AD\times AC+CD\times AC$

$=AC(AD+CD)$

$=AC\times AC$

$\therefore {AB}^2+{BC}^2={AC}^2$ $\left[henceproved\right]$